用php返回json数据 [英] Return json data with php
问题描述
我想返回json数据,但是我的代码无法正常工作.我没有收到任何错误消息.我有index.php,ajax.php和db.php. Db.php正在工作.但是我的ajax代码无法正常工作.我的错误在哪里?
I want to return json data but my code is not working. I don't get any error message. I have index.php, ajax.php and db.php. Db.php is working. But my ajax code is not working. Where is my mistake?
index.php:
<!DOCTYPE html>
<html>
<head>
<script type="text/javascript" src="http://code.jquery.com/jquery.min.js"></script>
</head>
<body>
<div id="test" style="width:500px;height:400px; margin:20px auto;"></div>
<script>
$(window).load(function () {
$.ajax({
dataType: "json",
url: 'ajax.php',
success:function(data){
$("#test").html(data);
}
});
});
</script>
</body>
</html>
Ajax.php:
<?php
require 'db.php';
$query="select lat,lng from locations order by id";
$result = pg_query($link, $query);
if (!$result) {
echo "An error occurred_xxx.\n";
}else {
$arr = pg_fetch_all($result);
echo json_encode($arr);
} ?>
推荐答案
如果您期望JSON
,则无论如何都要发送它.脚本错误时您正在做什么,正在发送text/html
.试试这个:
If you're expecting JSON
you need to send it regardless. What you're doing when your script errors, is sending text/html
. Try this:
header("Content-Type: application/json");
require 'db.php';
$query="select lat,lng from locations order by id";
$result = pg_query($link, $query);
$response = array();
if (!$result) {
$response = array(
'status' => false,
'message' => 'An error occured...'
);
}else {
$response = array(
'status' => true,
'message' => 'Success',
'data' => ph_fetch_all($result)
);
}
echo json_encode($response);
现在,如您所见,我们通过设置正确的Content-Type
标头而不是混合纯文本和json来发送实际的JSON.
Now as you'll see, we send actual JSON, by setting a correct Content-Type
header and not mixing plain text and json up.
要在jQuery中处理此响应,只需对响应进行条件处理:
To handle this response within your jQuery, simply condition the response:
$(window).load(function () {
$.ajax({
dataType: "json",
url: 'ajax.php',
success:function(data){
if(!data.status) {
$("#test").html("ERROR: " + data.message);
} else if(data.status) {
$("#test").html(data);
}
}
});
});
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