douxue7196 2018-01-17 18:26
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如何使用html id构建数组，然后在Javascript按钮onClick中将其作为JSON与ajax一起发送？

I am tying to use Post functions to send data from one php file to another, this is how my html code looks like (send.php):

    <!DOCTYPE html>
    <html>
    <head></head>
    <body>
    <p>operator:<input type="text" , id="operator"></p>
    <p>column:<input type="text" , id="columnName"></p>
    <button id="Btn">Click</button>
    <script src="js/jquery.min.js"></script>
    <script src="js/actions.js"></script>
    </body>
    </html>

This is how my JavaScript looks like (actions.js):

    $('#Btn').on('click', function () {
        var array = {'operator': $('#operator'), 'columnName': 
    $('#columnName')};
        $.ajax({
             url: 'receive.php',
             dataType: 'json',
             type: 'post',
             data: array,
             success: function (data) {
                if(data.success){
                     alert(data.operator);
                 } else {
                    alert("error");
                 }
             }

         });
     });

And this is what my receive.php code looks like:

    <?php
    header('Content-type: text/javascript');
    $json = array(
    'operator' => "",
    'columnName' => "",
     );
    if (isset($_POST['operator'],$_POST['columnName'])){
        $operator = $_POST['operator'];
        $columnName = $_POST['columnName'];

        $json['operator'] = $operator;
        $json['columnName'] = $columnName;
     }
     echo json_encode($json);
     ?>

When I fill both fields (column and operator) and I click on the button nothing happens! Any suggestions?

  • 写回答

1条回答 默认 最新

  • dongyan1993 2018-01-17 18:31
    关注

    firstly your html inputs are wrong it should be like 

    <p>operator:<input type="text" id="operator"></p>
<p>column:<input type="text" id="columnName"></p>

    should be separated by space only no comma(,).

    In JS use 

     var array = {'operator': $('#operator').val(), 'columnName': 
$('#columnName').val()};

    .val() to read the value of the input field

    also some changes in your php code the if() loop where you are checking for isset() only one input can be checked in isset() for another value you need to add && operator and another isset() to check for another input. here is the code

    if (isset($_POST['operator']) && isset($_POST['columnName'])){
    $operator = $_POST['operator'];
    $columnName = $_POST['columnName'];

    $json['operator'] = $operator;
    $json['columnName'] = $columnName;
    $json['success'] = TRUE;
 }
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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