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我正在将记录细节加载到允许用户编辑的模式中。我试图实现的是让用户更新模式中的记录，然后通过AJAX/jQuery提交给MySQL表，但是，在按下“保存更改”按钮后，什么都不会发生。我检查了JS Query并确认该按钮已正确链接，并在直接寻址PHP更新脚本时设法更新数据库。不知道为什么脚本拒绝启动jquery + AJAX - 通过Modal更新MySQL记录
模态：
<div id="output"></div>
<!-- Modal MYMODAL -->
<div class="modal fade" id="myModal" tabindex="-1" role="dialog" aria-labelledby="myModalLabel">
<div class="modal-dialog" role="document">
<div class="modal-content">
<div class="modal-header">
<button type="button" class="close" data-dismiss="modal" aria-label="Close"><span aria-hidden="true">×</span></button>
<h4 class="modal-title" id="myModalLabel">Edit Record</h4>
</div>
<div class="modal-body">
<!-- ID No. -->
<div class="form-group">
<label>ID No.:</label>
<input type="number" class="form-control" id="dataPID" name="dataPID" size="5" disabled />
</div>
<!-- /.id number -->
<!-- Category -->
<div class="form-group">
<label>Category:</label>
<input type="text" class="form-control" id="dataCat" name="dataCat" />
</div>
<!-- /.category -->
<!-- Issue -->
<div class="form-group">
<label>Issue:</label>
<input type="text" class="form-control" id="dataIssue" name="dataIssue" />
</div>
<!-- /.issue -->
<!-- Department Responsible -->
<div class="form-group">
<label>Department Responsible:</label>
<input type="text" class="form-control" id="dataDeptResp" name="dataDeptResp" />
</div>
<!-- /.department responsible -->
<!-- Experience -->
<div class="form-group">
<label>Experience:</label>
<input type="text" class="form-control" id="dataExp" name="dataExp" />
</div>
<!-- /.experience -->
<!-- textarea -->
<div class="form-group">
<label>Description:</label>
<textarea class="form-control" id="dataDesc" name="dataDesc" rows="3" ></textarea>
</div>
</div>
<div class="modal-footer">
<button type="button" id="SaveChanges" name="SaveChanges" class="btn btn-primary">Save Changes</button>
<button type="button" id="DeleteRecord" name="DeleteRecord" class="btn btn-danger">Delete Record</button>
<button type="button" class="btn btn-default" data-dismiss="modal">Close</button>
</div>
</div>
</div>
</div>
<!-- /.Modal MYMODAL -->
的Javascript：
$("#SaveChanges").click(function() {
$.ajax({
type: "POST",
url: "plugins/MySQL/ajax_action.php",
data: { action:"update_mysqli",PID:$("#dataPID").val(), Category:$("#dataCat").val(), Issue:$("#dataIssue").val(), Department_Responsible:$("#dataDeptResp").val(), Experience:$("#dataExp").val(), Description:$("#dataDesc").val()}, //your form data to post goes here as a json object
dataType: "json",
contentType:"text",
success: function(data) {
$('#output').html(data);
drawVisualization();
},
});
});
ajax_action.php
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
if(isset($_POST['action']) && ($_POST['action']=='update_mysqli')) {
// include connection details
include 'connect_db.php';
//Open a new connection to the MySQL server
$db = new mysqli($dbhost,$dbuser,$dbpass,$dbname);
//Output any connection error
if ($db->connect_error) {
die('Error : ('. $db->connect_errno .') '. $db->connect_error);
}
// get variables and sanitize
$pid = mysqli_real_escape_string($db,$_POST['PID']);
$cat = mysqli_real_escape_string($db,$_POST['Category']);
$issue = mysqli_real_escape_string($db,$_POST['Issue']);
$dept_resp = mysqli_real_escape_string($db,$_POST['Department_Responsible']);
$exp = mysqli_real_escape_string($db,$_POST['Experience']);
$desc = mysqli_real_escape_string($db,$_POST['Description']);
// check if record exists based on ID number
$result = $db->query("SELECT * FROM qci_problems_index_new WHERE PID='".$pid."'");
// if record is found, update accordingly
if ($result->num_rows > 0){
$sql = "UPDATE qci_problems_index_new SET Category = '$cat', Issue = '$issue', Department_Responsible = '$dept_resp', Experience = '$exp', Description = '$desc' WHERE PID = '$pid'";
if (!$db->query($sql)) {
echo "Error - Update of record PID " . $pid . " failed: (" . $db->errno . ") " . $db->error;
}
} else {
// if no record with relevant PID is found, create new record
$sql = "INSERT INTO `qci_problems_index_new`(`PID`, `Category`, `Issue`, `Department_Responsible`, `Experience`, `Description`) VALUES ('".$pid."', '".$cat."', '".$issue."', '".$dept_resp."', '".$exp."', '".$desc."')";
if (!$db->query($sql)) {
echo "Error - could not insert new record: (" . $db->errno . ") " . $db->error;
}
}
echo "Success, record updated successfully";
//close connection
$db->close();
}
编辑1： 铬控制台说以下内容： 
编辑2： 更新的代码
您是否检查了控制台的错误？检查了PHP错误日志中的错误？检查网络选项卡以查看是否已经发出实际呼叫，并且如果有，实际响应是什么？ –
Btw ..你打开[SQL Injections]（http://php.net/manual/en/security.database.sql-injection.php），并且应该使用[Prepared Statements]（http：// php。 net/manual/en/mysqli.quickstart.prepared-statements.php）而不是串联你的查询。 'mysqli_real_escape_string（）'不像人们想象的那么安全...... –
我没有检查，我的日志没有显示任何东西。该请求会通过并直接寻址PHP脚本时起作用。 – Armitage2k