我一直在尝试同时处理多个mySQL更新。我有4个选择/选项框，可以从数据库表中提取条目。我希望能够使用JQuery更新db onChange。我已经设法使这个工作与一个选择模块，但一旦我添加更多它旋转出来。我知道主要的错误代码在db_submit.php中，但真的不知道如何编写它。我知道必须有一个更干净的方法来做这件事。
表单页-INPUT.PHP
<html>
<head>
<script src="../assets/scripts/jquery-2.0.3.min.js"></script>
<script>
function updateDb() {
$.post("db_submit.php", $("#console").serialize());
}
</script>
<?php
include 'db_connect.php';
?>
</head>
<body>
<form id="console">
<select id="frame1" name="frame1" onChange="updateDb()">
<option value="">Select Channel</option>
<?php
$result = mysqli_query($con,"SELECT * FROM feedContent");
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['url'] . "'>" . $row['name'] . "</option>";
}
?>
</select>
<select id="frame2" name="frame2" onChange="updateDb()">
<option value="">Select Channel</option>
<?php
$result = mysqli_query($con,"SELECT * FROM feedContent");
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['url'] . "'>" . $row['name'] . "</option>";
}
?>
</select>
<select id="frame3" name="frame3" onChange="updateDb()">
<option value="">Select Channel</option>
<?php
$result = mysqli_query($con,"SELECT * FROM feedContent");
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['url'] . "'>" . $row['name'] . "</option>";
}
?>
</select>
<select id="frame4" name="frame4" onChange="updateDb()">
<option value="">Select Channel</option>
<?php
$result = mysqli_query($con,"SELECT * FROM feedContent");
while ($row = mysqli_fetch_array($result)) {
echo "<option value='" . $row['url'] . "'>" . $row['name'] . "</option>";
}
?>
</select>
</form>
</body>
<?php
mysqli_close($con);
?>
</html>
处理页-DB_SUBMIT.PHP
<?php
include 'db_connect.php';
$frame1= mysqli_escape_String($con,$_POST['frame1']);
$frame2= mysqli_escape_String($con,$_POST['frame2']);
$frame3= mysqli_escape_String($con,$_POST['frame3']);
$frame4= mysqli_escape_String($con,$_POST['frame4']);
$query = "UPDATE frameContent SET url='".$frame1."' WHERE name='frame1'";
$query = "UPDATE frameContent SET url='".$frame2."' WHERE name='frame2'";
$query = "UPDATE frameContent SET url='".$frame3."' WHERE name='frame3'";
$query = "UPDATE frameContent SET url='".$frame4."' WHERE name='frame4'";
mysqli_query($con,$query);
mysqli_close($con);
?>
我知道不断地设置$query变量会导致问题，但我不确定如何在一个页面中完成这项工作。任何帮助都将不胜感激。
谢谢！
最佳答案
首先确保$queries是连接的，然后用分号终止每个查询。之后，您可以使用mysqli_multi_query在一个php调用中执行所有四个更新。
$query = "UPDATE frameContent SET url='".$frame1."' WHERE name='frame1';";
$query .= "UPDATE frameContent SET url='".$frame2."' WHERE name='frame2';";
$query .= "UPDATE frameContent SET url='".$frame3."' WHERE name='frame3';";
$query .= "UPDATE frameContent SET url='".$frame4."' WHERE name='frame4';";
mysqli_multi_query($con,$query);