duanfen2349 2014-11-22 21:09
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在php中获取输入类型按钮的值

I wrote this code but the problem is that when I press the "Change background" button, nothing changes but I should be able to see some part of contents after I pressed it.

<form name="change "action="index.php" method="get">
        <center><a href="index.php"><button type="button">REFRESH THE PAGE!!</button></a></center><br/>
        <center><b>WELCOME NOTE!!</b></center><br/>
        <center><textarea readonly="readonly" name="textarea" rows="6" cols="50" style="color:blue; font-size:15pt">Each day holds a surprise. But only if we expect it can we see, hear, or feel it when it comes to us. Let's not be afraid to receive each day's surprise, whether it comes to us as sorrow or as joy It will open a new place in our hearts, a place where we can welcome new friends and celebrate more fully our shared humanity.</textarea></center>
<br/>

<?php

        mysql_connect("localhost","DB","password") or die("ERROR!!");
        mysql_select_db("DB") or die("COULDN'T FIND IT!!") or die("COULDN'T FIND DB"); 


        $sql = mysql_query("SELECT * FROM background");

        $id = 'ID';
        $Blue = 'blue';
        $White = 'white';
        $Silver = 'silver';
        $Red = 'red';
        $text=$_GET['textarea'];

        while($rows = mysql_fetch_assoc($sql)){


            if (isset( $_SESSION['CurrentUser'])){

            echo '<center><button type="button" name="background">Change background</button>';
            echo '<button type="button" name="color">Change font color</button>';
            echo '<button type="button" name="size">Change font size</button></center><br/>'; 

                if (isset( $_GET['background'])){ 
                echo '<span>Choose background color</span><br/>';
                echo '<a href="?colour='.$Blue.'"><img src="red.png"></a>'; 
                echo '<a href="?colour='.$White.'"><img src="white.jpg"></a>'; 
                echo '<a href="?colour='.$Silver.'"><img src="silver.jpg"></a>';
                echo '<a href="?colour='.$Red.'"><img src="red.png"></a>'; }

            }               

    }
    ?>

    </form>


    </td></tr></table>

Program doesn't see this part;

if (isset( $_GET['background'])){ 
                    echo '<span>Choose background color</span><br/>';
                    echo '<a href="?colour='.$Blue.'"><img src="red.png"></a>'; 
                    echo '<a href="?colour='.$White.'"><img src="white.jpg"></a>'; 
                    echo '<a href="?colour='.$Silver.'"><img src="silver.jpg"></a>';
                    echo '<a href="?colour='.$Red.'"><img src="red.png"></a>'; }
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3条回答 默认 最新

  • dtng25909 2014-11-22 21:31
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    It doesn't work because form cannot be submited without submit button: Replace <button type="button" with <button type="submit"

    BTW use mysqli_ instead of mysql_ because it is deprecated

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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