JQuery post to php
Why is it that I can never get JQuery serialize to work properly. I guess I must be missing something.
I can serialize a form data an开发者_JS百科d it shows in an alert:
var forminfo = $j('#frmuserinfo').serialize();
alert(forminfo);
I then post to my PHP page thus:
$j.ajax({
type: "POST",
url: "cv-user-process.php",
data: "forminfo="+forminfo,
cache: false,
complete: function(data) {
}
});
But WHENEVER (not the first time) I try to insert/update the data in the DB I only ever get 1 varaible passed:
Here is my PHP script:
$testit = mysql_query("UPDATE cv_usersmeta SET inputtest='".$_POST['forminfo']."' WHERE user='X'");
the data passed only ever gets the first variable. why?
I think it is more the way I deal with the php but it drives me nuts and always takes me far too long to find where I am going wrong.
I believe it's because you're wrapping the entire serialized form under the forminfo variable.
This would result in a form var such as forminfo=field1=val1&field2=val2
- so the whole thing comes to php as forminfo
You don't need to do that:
$j.ajax({
type: "POST",
url: "cv-user-process.php",
data: forminfo,
cache: false,
complete: function(data) {
}
In your php script have this at the top:
<?php
print_r($_POST);
exit;
?>
This should help you in finding out what could be going wrong for you.
You need to pass it a little differently, like this:
$j.ajax({
type: "POST",
url: "cv-user-process.php",
data: $j('#frmuserinfo').serialize(),
cache: false,
complete: function(data) {
}
});
In your php don't use $_POST['forminfo']
but rather $_POST['nameOfTheInput']
. Further, you should sanitize the input, like this:
$testit = mysql_query("UPDATE cv_usersmeta SET inputtest='".mysql_real_escape_string($_POST['nameOfTheInput'])."' WHERE user='X'");
This prevents SQL injection attacks, which I suggest you read more about :)
I'd suggest you to ajaxForm/ajaxSubmit from http://jquery.malsup.com/form/ on the client side.
And if you still need to send the whole serialized data, you must encode your string with escape()
function. Like this:
$j.ajax({
data: "forminfo="+escape(forminfo),
...
});
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