我想在数据插入之前在PHP中使用Java脚本警报。我能够将数据插入数据库,但echo无法完成其工作。这可能只是一个简单的错误,但我无法发现它:(
/// Java script
/// send data for insertion
$.get("tute7_add.php",{unit_code:unit_code,unit_name:unit_name,lecturer:lecturer,semester:semester}).done(function(data){
alert('You have added a unit successfully!');
});
tute7_add.php
<?php
// I Can't send alert!!
echo "<script type='text/javascript'>
alert('Don\'t have a record');
</script>";
$code= $_GET['unit_code'];
$name= $_GET['unit_name'];
$lecturer= $_GET['lecturer'];
$semester= $_GET['semester'];
$query = "INSERT INTO units (`unit_code`,`unit_name`,`lecturer`,`semester`) VALUES ('$code','$name','$lecturer','$semester');";
$mysqli->query($query);
?>
您正在使用Ajax请求URL。
The
data
variable will have a chunk of HTML, including the script element, in it.You aren't doing anything with the value of
data
. You're just ignoring it.通常,在处理Ajax时,您将以JSON之类的格式而不是HTML返回原始数据。然后,您可以使用客户端代码进行处理。
例如
随着: