PHP 将 JSON 返回给 JQUERY AJAX CALL
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PHP returning JSON to JQUERY AJAX CALL
提问by mcl
I am still struggling to get my head around the ins and out of JQUERY, AJAX and PHP.
我仍在努力了解 JQUERY、AJAX 和 PHP 的来龙去脉。
I can now call the PHP OK, process the form elements and send an email, but I am not handling the return to the AJAX. I am always getting the error:selector activated and when I try to list the supposed JSON returned, I get info, that is obviously wrong.
我现在可以调用 PHP OK，处理表单元素并发送电子邮件，但我没有处理对 AJAX 的返回。我总是error:激活选择器，当我尝试列出返回的假定 JSON 时，我得到信息，这显然是错误的。
PHP with supposed JSON return
PHP 与假定的 JSON 返回
<?php
touch('phpTouch.txt');
// process email
$email=1;
if ($email) {
$value = array('return' => 1, 'msg1' => 'Message sent OK, we will be in touch ASAP');
} else {
$value = array('return' => 0, 'msg1' => 'Message Failed, please try later');
}
$output = $json->encode($value);
echo $output;
?>
Javascript and AJAX
Javascript 和 AJAX
function submitForm(evt) {
$('#msgid').html('<h1>Submitting Form (External Routine)</h1>');
if ($('#formEnquiry').valid() ) {
$("#msgid").append("<h1>(Outside Ready) VALIDATED send to PHP</h1>");
$.ajax({
url: "ContactFormProcess3.php",
type: "POST",
data: $('#formEnquiry').serialize(),
dataType: "json",
success: function (data) {
alert("SUCCESS:");
for(var key in data) {
$('#msgid').append(key);
$('#msgid').append('=' + data[key] + '<br />');
}
},
error: function (data) {
alert("ERROR: ");
for(var key in data) {
$('#msgid').append(key);
$('#msgid').append('=' + data[key] + '<br />');
}
}
});
} else {
$('#msgid').append('<h1>(Outside Ready) NOT VALIDATED</h1>');
}
evt.preventDefault();
};
Listing of supposed JSON data
假设的 JSON 数据列表
readyState=4
setRequestHeader=function (a,b){if(!s){var c=a.toLowerCase();a=m[c]=m[c]||a,l[a]=b}return this}
getAllResponseHeaders=function (){return s===2?n:null}
getResponseHeader=function (a){var c;if(s===2){if(!o){o={};while(c=bF.exec(n))o[c[1].toLowerCase()]=c[2]}c=o[a.toLowerCase()]}return c===b?null:c}
overrideMimeType=function (a){s||(d.mimeType=a);return this}
etc etc
If anyone can advise as to what stupid mistake I have made, then I would be most grateful.
如果有人能告诉我我犯了什么愚蠢的错误，我将不胜感激。
回答by Arnaud Le Blanc
You can return json in PHP this way:
您可以通过这种方式在 PHP 中返回 json：
header('Content-Type: application/json');
echo json_encode(array('foo' => 'bar'));
exit;
