Question

我目前正在使用$ _POST函数导入数据。我想使用导入的值作为选定列。 Currenlty我的代码如下：

$value = $_POST["name"]

$sql = "SELECT question_id, question FROM Question WHERE question_id= $value";

当我使用JQuery调用我的php脚本时，我当前收到错误。代码的这一部分是潜在的来源吗？

这是我正在使用的JQuery代码。

<script type="text/javascript">
    $(document).ready(function(){
        $('button').click(function(){
            $('#show').load('/php/getQuestion.php'), { name: $("#txtname").val()}
        });

    });
</script>

Txtname为<strong>Name:</strong><input type = "text" id = "txtname">

错误是

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Answer 1

尝试

$sql = "SELECT question_id, question FROM Question WHERE question_id = ".$value;

希望有所帮助

Answer 2

你的$.load()关闭是不正确的，它需要第二个参数作为参数，但你在传递你的参数之前关闭它

$(document).ready(function(){
    $('button').click(function(){
        $('#show').load('/php/getQuestion.php',
          { name: $("#txtname").val()} );
               // ---------------------^ closing parenthesis of load()
    });
});

在PHP中，您在$value = $_POST["name"]

之后错过了分号

Answer 3

首先将您的参数包装为字符串

$value = $_POST["name"];
$sql = "SELECT `question_id`, `question` FROM Question WHERE `question_id` like '%$value%'";

或

$value = (integer)$_POST["name"];
$sql = "SELECT `question_id`, `question` FROM Question WHERE `question_id` = {$value}";

使用带有$ _POST的WHERE函数

3 个答案: