html上如何在一个button的右上方加一个可以删除该button的x,并且可以有alert确认是否删除

想要做到的样式

图片说明

当鼠标放在button上时,出现关闭按钮,点击关闭按钮时,弹出提示

图片说明

前端小白,跪求大神解答。。。。

4个回答

按钮需要写在div或者其他容器标签里面,添加相对定位,关闭也是一个按钮或者添加onclick的其他标签,加上绝对定位,类似这样:

<div id="aaa" style="float: left; position:relative;">
    <input type="button" value="按钮" />
    <input type="button" value="×" style="position: absolute;">
</div>

关闭的那个按钮样式写成display:none; 另加个 父级div:hover ×按钮{display: block;} 样式,
然后给那个删除按钮加点击事件,大体方法内容:

 var state = confirm("是否删除?");
 if(state){
 $("#aaa").remove();
 }else{
 return false;}
Skr_Eric
Skr-Eric 感谢解决思路,虽然我不是这么解决的。。。
10 个月之前 回复

按钮用div实现,删除图标用相对定位。

f12打开控制台,找到那个x,看它是否有什么id,或者唯一的name或class属性,然后使用

 $(document).on("click", "#id,name,或者class属性", function () {
                         confirm("是否删除?");
        });

哪个×可以用after伪元素 定位到button 的右上角,点击×获取到button 的dom 然后confirm判断一下 确认就remove() button

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如题, 我想在一个Chart里面画出三个波形图,就将三个波形的数据都放在一个数组myDatasets里面。然后再赋值给Chart,但运行的时候发现整个循环里面所有的东西都没运行,包括 ``` console.log ``` 也没运行,但控制台又没报错,实在一头雾水,是不是我语法哪里有问题?求大佬帮帮忙QAQ 这是我整个函数↓ ``` function drawWaveform(datas){ console.log("绘制开始。。。"); var lineChart = $("#waveform"); console.log(datas); var xs=[]; var allYs=[]; var myDatasets=[]; var ys=[]; var k=0; for(;k<datas.length;k++){ //【从这里开始就没运行了】 console.log(k); var points=datas[k].points; console.log(points); for(var i = 0;i<points.length;i++){ if(i==0){ xs.push(points[i].x); } ys.push(points[i].y); } allYs.push(ys); var color=""; if(datas[k].phrase=="A"){ color="#FFD700";//gold(yellow) var curDataset={ label: datas[k].phrase, fill: false, lineTension: 0.1, backgroundColor: "#fff", borderColor: color, borderCapStyle: 'butt', borderDash: [], borderDashOffset: 0.0, borderJoinStyle: 'miter', pointBorderColor: "#fff",//white pointBackgroundColor: "#141E41",//black pointBorderWidth: 1, pointHoverRadius: 10, pointHoverBackgroundColor: "#FC2055",//red pointHoverBorderColor: "#fff", pointHoverBorderWidth: 1, pointRadius: 0, pointHitRadius: 10, data: ys, spanGaps: false }; this.myDatasets.push(curDataset); } if(datas[k].phrase=="B"){ color="#047bf8";//blue var curDataset={ label: datas[k].phrase, fill: false, lineTension: 0.1, backgroundColor: "#fff", borderColor: color, borderCapStyle: 'butt', borderDash: [], borderDashOffset: 0.0, borderJoinStyle: 'miter', pointBorderColor: "#fff",//white pointBackgroundColor: "#141E41",//black pointBorderWidth: 1, pointHoverRadius: 10, pointHoverBackgroundColor: "#FC2055",//red pointHoverBorderColor: "#fff", pointHoverBorderWidth: 1, pointRadius: 0, pointHitRadius: 10, data: ys, spanGaps: false }; this.myDatasets.push(curDataset); } if(datas[k].phrase=="C"){ color="#FC2055";//red var curDataset={ label: datas[k].phrase, fill: false, lineTension: 0.1, backgroundColor: "#fff", borderColor: color, borderCapStyle: 'butt', borderDash: [], borderDashOffset: 0.0, borderJoinStyle: 'miter', pointBorderColor: "#fff",//white pointBackgroundColor: "#141E41",//black pointBorderWidth: 1, pointHoverRadius: 10, pointHoverBackgroundColor: "#FC2055",//red pointHoverBorderColor: "#fff", pointHoverBorderWidth: 1, pointRadius: 0, pointHitRadius: 10, data: ys, spanGaps: false }; this.myDatasets.push(curDataset); } ys=[]; //【到这里结束】 } //【这里开始有结果】 console.log(myDatasets); // line chart data var lineData = { labels: xs, datasets: myDatasets }; // line chart init var myLineChart = new Chart(lineChart, { type: 'line', data: lineData, options: { legend: { display: false }, scales: { xAxes: [{ ticks: { fontSize: '11', fontColor: '#969da5' }, gridLines: { color: 'rgba(0,0,0,0.05)', zeroLineColor: 'rgba(0,0,0,0.05)' } }], yAxes: [{ display: false, ticks: { beginAtZero: true, max: 65 } }] } } }); console.log("绘制结束"); } ``` 在这个函数的ajax里面调用的这个函数↓ ``` function show3(id1,id2,id3) { console.log("show3"); var ids=[id1,id2,id3]; var flag = false; //标识,表示页面上数据还未处理完成 var datas=[]; console.log(ids); for(var i=0;i<3;i++){ //console.log(ids); var curid=ids[i]; $.ajax({ type : "get", url : "observe/getImage", data : { ID : curid }, beforeSend: function () { ShowDiv(); }, complete: function () { HiddenDiv(); }, success : function(data) { console.log(data) datas.push(data); } }); } console.log(datas); //绘制波形图 //drawWaveform(data.points); drawWaveform(datas);//【**在这里调用的**】 $("#closeDivBtn").html("<input class=\"mr-2 mb-2 btn btn-primary btn-lg\" type=\"button\" value=\"关闭\" onclick=\"CloseDiv('MyDiv','fade')\">"); AlertDiv('MyDiv','fade'); } ```

javascript中的onclick功能按钮无法正常工作

<div class="post-text" itemprop="text"> <p>I am a newbie in jquery, I have a stupid question for onclick function. Please help me. I have a table, when viewer click "Add new" button, I will using jquery add a row to existing table. In new row, there are 2 textbox to input data and a button to transfer data from textbox to php file. This is my code: html code:</p> <pre><code>&lt;INPUT type="button" value="Add Row" onclick="addRow('dataTable')" /&gt; &lt;INPUT type="button" value="Delete Row" onclick="deleteRow('dataTable')" /&gt; &lt;TABLE id="dataTable" width="350px" border="1"&gt; &lt;thead&gt; &lt;td&gt;No&lt;/td&gt; &lt;td&gt;First name&lt;/td&gt; &lt;td&gt;Last name&lt;/td&gt; &lt;/thead&gt; &lt;?php include("functions/connect_db.php"); $no=0; $sql="SELECT * FROM `fullnames`"; $result=mysqli_query($conn,$sql) or die('Could not select Machine'.mysqli_error($conn)); while ($set=mysqli_fetch_array($result,MYSQLI_ASSOC)) { $no++; $first=$set['firstname']; $last=$set['lastname']; ?&gt; &lt;tbody&gt; &lt;tr&gt; &lt;td&gt;&lt;?php echo $no; ?&gt;&lt;/td&gt; &lt;td&gt;&lt;?php echo $first; ?&gt;&lt;/td&gt; &lt;td&gt;&lt;?php echo $last; ?&gt;&lt;/td&gt; &lt;/tr&gt; &lt;?php } ?&gt; &lt;/tbody&gt; &lt;/TABLE&gt; </code></pre> <p>script:</p> <pre><code>function addRow(tableID) { var table = document.getElementById(tableID); var rowCount = table.rows.length; var row = table.insertRow(rowCount); var cell1 = row.insertCell(0); cell1.innerHTML = rowCount ; var cell2 = row.insertCell(1); var element1 = document.createElement("input"); element1.type = "text"; element1.id="first_input_"+rowCount; cell2.appendChild(element1); var cell3 = row.insertCell(2); var element2 = document.createElement("input"); element2.type = "text"; element2.id="last_input_"+rowCount; cell3.appendChild(element2); var cell4=row.insertCell(3); var element3 = document.createElement('input'); element3.setAttribute('type','button'); element3.id="#button_"+rowCount; element3.className="butt_class"; element3.value='Save'; cell4.appendChild(element3); } $(document).ready(function() { $(".butt_class").click(function() { var ID=$(this).attr('id'); var first=$("#first_input_"+ID).val(); var last=$("#last_input_"+ID).val(); if(first.length&gt;0 &amp;&amp; last.length&gt;0) { $.ajax({ type: "POST", url: "add_new.php", data: dataString, cache: false }); } }) }) </code></pre> <p>When I click to Save button in each row, nothing is happen. I tried by replace all click Save function with alert but there is not change. Pls help me. Thank you in advance!</p> </div>

Sendmail PHP无法正常工作 - Go Daddy Hosting

<div class="post-text" itemprop="text"> <p>I think this is straightforward, but for some reason it's not working. Does anyone see a problem with my code? If not, I'm guessing it's the hosting that's the issue.</p> <p>Here's the code:</p> <p>HTML</p> <pre><code>&lt;form action="sendmail.php" name="form1" method="post"&gt; &lt;input name="hidSubmit" type="hidden" value="true" id="hidSubmit" /&gt; &lt;input type="text" placeholder="Your Name" name="txtName" required&gt;&lt;br&gt; &lt;input type="email" placeholder="Email Address" name="txtEmail" required&gt;&lt;br&gt; &lt;input type="number" placeholder="Phone Number" name="txtPhone"&gt;&lt;br&gt; &lt;button type="submit" value="Submit" class="butt"&gt;&lt;strong&gt;Submit&lt;/strong&gt;&lt;/button&gt; &lt;/form&gt; </code></pre> <p>PHP</p> <pre><code>&lt;?php //Check whether the submission is made if(isset($_POST["hidSubmit"])){ //msg body $txtMsg = "Enquiry From Website Name : ".$_POST["txtName"]." Email: ".$_POST["txtEmail"]." Phone: ".$_POST["txtPhone"]." "; //Declarate the necessary variables $mail_to = "info@insidemarketstrategy.com"; $mail_from= "admin@insidemarketstrategy.com"; $mail_sub="Inquiry from insidemarketstrategy.com Website"; $mail_mesg=$txtMsg; $headers = "From: admin@insidemarketstrategy.com " . "X-Mailer: php" . "Content-type: text/html; charset=iso-8859-1 "; //Check for success/failure of delivery if(mail($mail_to,$mail_sub,$mail_mesg,$headers)) echo "&lt;div class='alert alert-success alert-dismissable' id='alertmsg'&gt; &lt;button type='button' class='close' data-dismiss='alert' aria-hidden='true'&gt;&amp;times;&lt;/button&gt; &lt;h3&gt;Thank you !&lt;br&gt;&lt;br&gt;&lt;span&gt;we will get in touch with you soon&lt;/span&gt;&lt;/h3&gt; &lt;/div&gt;"; else echo "&lt;div class='alert alert-danger alert-dismissable'&gt;&lt;center&gt;&lt;span class='textred' align='center'&gt;&lt;b&gt;Failed to send the E-mail from $mail_sub to $mail_to&lt;/b&gt;&lt;/span&gt;&lt;/center&gt;&lt;/div&gt;"; } echo ("&lt;meta http-equiv='refresh' content='5 ;url=index.html'&gt;"); ?&gt; </code></pre> </div>

Ajax和jQuery与PHP

<div class="post-text" itemprop="text"> <p>I have this code which posts data using jQuery and ajax to itself.</p> <pre><code>&lt;input type="button" id="butt" value="button11111111111111" &gt; &lt;script&gt; $("#butt").on('click',function(e) { $.ajax( { type:'POST', url:'test.php', data:product_type:"cake" }); }); &lt;/script&gt; &lt;?php if(isset($_POST['product_type'])) $abc=$_POST['product_type']; if(isset($abc)) echo $abc; ?&gt; </code></pre> <p>Now when I try to run this code.I do get the ok status inside the <strong>Network Console</strong> of Chrome but this code doesn't echo the output. </p> <p>I simply want to display the result that has been passed by the ajax method.</p> <p>I am new to ajax and jquery so i don't know much about how they work exactly but is it possible? If yes, then how could i achieve that without actually refreshing the page?</p> </div>

如何在jquery中重复多次的div部分中显示相应的名称?

<div class="post-text" itemprop="text"> <p>I have written a code in jquery for displaying a particular <code>&lt;div&gt;</code> section multiple times which contains format for displaying names.</p> <p>First I have a form for entering the names and age as shown below :</p> <pre><code>&lt;form action="" method="post" name="person" id="addperson"&gt; &lt;div&gt; &lt;div class="form-group col-md-12"&gt; &lt;label for="exampleInputEmail1"&gt;First Name&lt;/label&gt; &lt;input type="text" class="form-control" name="fname" id="fnname" placeholder="First Name" required&gt; &lt;div class="clearfix"&gt;&lt;/div&gt; &lt;/div&gt; &lt;div class="form-group col-md-12"&gt; &lt;label for="exampleInputEmail1"&gt;Last Name&lt;/label&gt; &lt;input type="text" class="form-control" name="lname" id="lnname" placeholder="Last Name" required&gt; &lt;div class="clearfix"&gt;&lt;/div&gt; &lt;/div&gt; &lt;div class="form-group col-md-7"&gt; &lt;label for="exampleInputEmail1" class="ctred"&gt;Age only if under 18&lt;/label&gt; &lt;input type="text" class="form-control" name="age" id="aage" placeholder="Age" required&gt; &lt;div class="clearfix"&gt;&lt;/div&gt; &lt;/div&gt; &lt;button type="submit" id="senddata" name="send" class="btn btn-primary mag_top2"&gt;Submit&lt;/button&gt; &lt;/div&gt; &lt;/form&gt; </code></pre> <p>The above code is for entering the names and age when clicking the submit button , it should enter the values into the database. For each values of names and age, a <code>&lt;div&gt;</code> section is displayed with the corresponding names entered. The div section is shown below in the <code>#pern</code> appended:</p> <pre><code>$(document).ready(function(){ $("#senddata").click(function(e) { $.ajax({ type : "POST", url : "data2.php", data : dataString, cache : false, success : function(result1){ var fname = firstname; var lname = lastname; var fullname = fname.concat(" ",lname); document.getElementById("fullname").innerHTML = fullname; //$("#fullname").html(fullname); } }); $('#addperson').trigger("reset"); $("#pern").append( "&lt;div class='pad pen col-xs-12 col-sm-12 col-md-12 pern alert' id='bloc_style'&gt;\ &lt;div class='pad col-xs-12 col-md-6'&gt;\ &lt;h4 id='fullname'&gt;&lt;/h4&gt;\ &lt;/div&gt;\ &lt;div class='pad col-md-5'&gt;\ &lt;span class='next-step'&gt;\ &lt;button class='ret_but butt label label-primary' id='equipment' name='equipment' type='button'&gt;Select Equipment&lt;/button&gt;\ &lt;/span&gt;\ &lt;div class='status'&gt;\ &lt;b&gt;Status&lt;/b&gt;&lt;i class='open' id='open'&gt;Open&lt;/i&gt;\ &lt;/div&gt;\ &lt;/div&gt;\ &lt;div class='pad col-md-1'&gt;\ &lt;button aria-hidden='true' data-dismiss='alert' class='closee' id='garbage' type='button'&gt; \ &lt;i class='fa fa-trash-o' aria-hidden='true'&gt;&lt;/i&gt;\ &lt;/button&gt;\ &lt;/div&gt;\ &lt;/div&gt;"); }); }); </code></pre> <p>For each names which I have entered in the form, it should display in the div given in <code>#pern</code> with the name. But with this code, for each name entered and clicking the submit button a div section is created but the name entered will be displayed in the previous div section displayed replacing the previous name entered. How to avoid this ? I want to display the name in the corresponding box only not replacing the previous box with the previous name.</p> </div>

Yii CgridView批量删除

<div class="post-text" itemprop="text"> <p>In many of my Models' CgridViews I have a bulk delete function: a chechboxColumn and a delete button which deletes all the checked users. For that I am using ajax in the admin and a new action in the controller.</p> <p>All this works fine until I add pagination to th gridview, which is not saving the checked rows in the previous pages.</p> <p>I tried to use 'enableHistory'=true, but it did nothing (and from what I'v read I'm not the only one :mellow: ) , so I downloaded this extension: selgridview</p> <p>The extension works - when I move through the pages , the checked rows stay checked BUT , my bulk delete function is seeing only the checked rows of the page I'm in right now.</p> <p>this is the ajax I'm using: </p> <pre><code> &lt;?php Yii::app()-&gt;clientScript-&gt;registerScript('delete',' $("#butt").click(function(){ var checked=$("#person-grid").yiiGridView("getChecked","person-grid_c11"); var count=checked.length; if(count&gt;0 &amp;&amp; confirm(" are you sure you want to delete "+count+" people ? ")) { $.ajax({ data:{checked:checked}, url:"'.CHtml::normalizeUrl(array('person/remove')).'", success:function(data){$("#person-grid").yiiGridView("update",{});}, }); } }); '); ?&gt; </code></pre> <p>Now , maybe thats a silly question but I know little about javascript. I'm not even sure that the problem is in the ajax . . . . </p> <p>Help would be much appreciated :rolleyes: </p> </div>

为在php中从数据库检索的数据创建一个后退按钮

<div class="post-text" itemprop="text"> <p>i have created a website for quiz exam. I have the following html and ajax for calling php functions in the file. i am calling questions from the database. 1 question at a time is shown to the user. there is no submit button, instead when the user clicks the option,the next question is shown.</p> <p></p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false"> <div class="snippet-code"> <pre class="snippet-code-js lang-js prettyprint-override"><code>&lt;script&gt; $(document).ready(function(){ $('body').on('click','input[type="radio"]', function(){ var curr_id = $('.question').data('nextQuestion'); var answer1 = $('#radio1').is(':checked'); var answer2 = $('#radio2').is(':checked'); var answer3 = $('#radio3').is(':checked'); var answer4 = $('#radio4').is(':checked'); getCorrectAnswer(curr_id, answer1, answer2, answer3, answer4); setTimeout(getQuestion.bind(this,curr_id, answer1, answer2, answer3, answer4), 1000); }); function getQuestion(curr_id, answer1=false, answer2=false, answer3=false, answer4=false){ console.log(curr_id); $.post("mohajax.php", { next_id: parseInt(curr_id)+1, answer1: answer1, answer2: answer2, answer3: answer3, answer4: answer4, }, function(data, status){ $('#container_for_questions').html(data); }); } function getCorrectAnswer(curr_id, answer1=false, answer2=false, answer3=false, answer4=false){ $.post("mohajax_get_correct_answer.php", { next_id: parseInt(curr_id), answer1: answer1, answer2: answer2, answer3: answer3, answer4: answer4, }, function(data, status){ $('#container_for_questions').html(data); }); } getQuestion(-1); }); &lt;/script&gt;</code></pre> <pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;div class="services"&gt; &lt;div class="zubi"&gt;&lt;div class="container" id="container_for_questions"&gt;&lt;/div&gt;&lt;/div&gt; &lt;/div&gt;</code></pre> </div> </div> <p>below is my php files for calling the questions</p> <p></p><div class="snippet" data-lang="js" data-hide="false" data-console="true" data-babel="false"> <div class="snippet-code"> <pre class="snippet-code-html lang-html prettyprint-override"><code>&lt;?php // Start the session session_start(); $con=mysqli_connect("localhost","root","","quiz"); // change here to your data // Check connection if (mysqli_connect_errno()) { echo "Failed to connect to MySQL: " . mysqli_connect_error(); } // Check the number of all questions, if next_id is more than last question, back to first or whatever you want; $response=mysqli_query($con,"select * from moh limit 25"); $number_of_all_questions = mysqli_num_rows($response); if($_POST['next_id'] == 0){ // reset to default $_SESSION["correct_score"] = 0; $_SESSION["not_correct_score"] = 0; } if($number_of_all_questions &lt;= $_POST['next_id']){ // Quiz finished, show results echo"&lt;div&gt; &lt;h2&gt;Results:&lt;/h2&gt; &lt;p&gt;Correct answers: {$_SESSION['correct_score']}&lt;/p&gt; &lt;p&gt;Wrong answers: {$_SESSION['not_correct_score']}&lt;/p&gt; &lt;/div&gt;"; }else{ // query next question $response=mysqli_query($con,"select * from moh WHERE id =(select min(id) from moh where id &gt; {$_POST['next_id']})"); ?&gt; &lt;?php while($result=mysqli_fetch_array($response,MYSQLI_ASSOC)){ ?&gt; &lt;div id="question_&lt;?= $result['id'] ?&gt;" class='question' data-next-question="&lt;?= $_POST['next_id'] ?&gt;"&gt; &lt;!--check the class for plurals if error occurs--&gt; &lt;h2&gt;&lt;?= $result['id'].".".$result['question_name'] ?&gt;&lt;/h2&gt; &lt;div class='align'&gt; &lt;input type="radio" value="1" id='radio1' name='1'&gt; &lt;label id='ans1' for='radio1'&gt;&lt;?= $result['answer1'] ?&gt;&lt;/label&gt; &lt;br/&gt; &lt;input type="radio" value="2" id='radio2' name='2'&gt; &lt;label id='ans2' for='radio2'&gt;&lt;?= $result['answer2'] ?&gt;&lt;/label&gt; &lt;br/&gt; &lt;input type="radio" value="3" id='radio3' name='3'&gt; &lt;label id='ans3' for='radio3'&gt;&lt;?= $result['answer3'] ?&gt;&lt;/label&gt; &lt;br/&gt; &lt;input type="radio" value="4" id='radio4' name='4'&gt; &lt;label id='ans4' for='radio4'&gt;&lt;?= $result['answer4'] ?&gt;&lt;/label&gt; &lt;/div&gt; &lt;br/&gt; &lt;?php /*&lt;input type="button" data-next-question="&lt;?= $_POST['next_id'] ?&gt;" id='next' value='Next!' name='question' class='butt'/&gt; */?&gt; &lt;/div&gt; &lt;?php }?&gt; &lt;?php }?&gt; &lt;?php mysqli_close($con); ?&gt;</code></pre> </div> </div> <p>i need a back button to go to the previous question. how can i do this.</p> </div>

在中国程序员是青春饭吗?

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程序员请照顾好自己,周末病魔差点一套带走我。

程序员在一个周末的时间,得了重病,差点当场去世,还好及时挽救回来了。

Java基础知识面试题(2020最新版)

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