javascript - 下拉 Onchange 并将 JavaScript 值传递给 PHP
问题描述
您好,我有一个关于如何将 JavaScript 值发送到 PHP 的问题。
这是我的表格:
来自数据库的下拉问题:
<form action="action/survey">
<div class="form-group">
<label class="control-label col-sm-2" for="email">Select Question</label>
<div class="col-sm-10">
<select class="form-control" id="mySelect" onchange="option()">
<?php
$sql = mysqli_query($con,"SELECT *,(SELECT GROUP_CONCAT(answer) AS `option`FROM `survey_anweroptions` WHERE survey_qID = sq.survey_qID) `option` FROM `survey_questionnaire`sq WHERE sq.survey_ID = $id");
while ($q = mysqli_fetch_array($sql)) {
?>
<option value="<?php echo $q[0]?>"><?php echo $q[2]?></option>
<?php
}
?>
</select>
</div>
</div>
<div class="form-group">
<label class="control-label " for=""></label>
</div>
<?php
$sql = mysqli_query($con,"SELECT *,(SELECT GROUP_CONCAT(answer) AS `option`FROM `survey_anweroptions` WHERE survey_qID = sq.survey_qID) `option` FROM `survey_questionnaire`sq WHERE sq.survey_ID = 1 AND survey_qID = 1");
$d1= mysqli_fetch_array($sql);
?>
<div class="form-group">
<label class="control-label col-sm-2" for="">Question</label>
<div class="col-sm-10">
<input type="text" class="form-control" id="" placeholder="" value="<?php echo $d1[2]?>">
</div>
</div>
<?php
$variable = $d1[3];
$z = 1;
$piece = explode(",", $variable);
foreach ($piece as $key => $value) {
?>
<div class="form-group">
<label class="control-label col-sm-2" for="">Option <?php echo $z?> </label>
<div class="col-sm-10">
<input type="text" class="form-control" id="" placeholder="" value="<?php echo $value?>">
</div>
</div>
<?php
$z++;
}
?>
<div class="text-center">
<button type="submit" class="btn btn-default" name="">Update</button>
</div>
</form>
JavaScript:
function option(){
var x = document.getElementById("mySelect").value;
document.getElementById("demo").innerHTML = "You selected: " + x;
}
我想要的是,如果我在下拉列表中选择 1 个值,则选项的值必须发送到此查询中的 PHP 变量:
SELECT *, (
SELECT GROUP_CONCAT(answer) AS `option`
FROM `survey_anweroptions`
WHERE survey_qID = sq.survey_qID
) `option`
FROM `survey_questionnaire`sq
WHERE sq.survey_ID = 1
AND survey_qID = (**the value of option will go here**)
解决方案
您应该使用名称而不是 id。并执行 print_r($_POST) 检查天气发布数据是否到来。在您使用 ajax 调用您的 php 脚本之前,您也不需要 javascript
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