[英]Getting JSON data out of PHP using .get() in jQuery
我正在尝试从$.get()
jQuery调用中获取JSON数据,但似乎无法使其正常运行。
这是我的代码:
var url = "getDetailsJSON.php?ImageID=" + escape(itemName);
$.get(url, function (data) {
console.log("Success!");
var $detailDiv = $("#description");
var itemDetails = $.parseJSON(data); // Is this how I would get the JSON object in the php code?
console.log("success" + data);
var children = $detailDiv.children();
for (var i = children.length; i > 0; i--) {
$detailDiv.remove(children[i - 1]);
}
var descriptionP = $("<p></p>");
descriptionP.text("Description: " + itemDetails.description);
$detailDiv.append(descriptionP);
var priceP = $("<p></p>");
priceP.text("Price: $" + itemDetails.price);
$detailDiv.append(priceP);
var list = $("<ul></ul>");
$.each(itemDetails.urls, function (index, value) {
var url = itemDetails.urls[index];
var li = $("<li></li>");
var a = $("<a></a>");
a.attr("href", url);
a.text(url);
li.append(a);
list.append(li);
});
$detailDiv.append(list);
});
这是PHP代码:
<?php
require_once('JSON.php');
$json = new Services_JSON();
$itemGuitar = array(
'id' => 'itemGuitar',
'description' => 'Pete Townshend once played this guitar while his own axe was in the shop having bits of drumkit removed from it.',
'price' => 5695.99,
'urls' => array('http://www.thewho.com/',
'http://en.wikipedia.org/wiki/Pete_Townshend')
);
$itemShades = array(
'id' => 'itemShades',
'description' => 'Yoko Ono\'s sunglasses. While perhaps not valued much by Beatles fans, this pair is rumored to have been licked by John Lennon.',
'price' => 258.99,
'urls' => array('http://www.beatles.com/',
'http://johnlennon.com/',
'http://www.yoko-ono.com/')
);
$itemCowbell = array(
'id' => 'itemCowbell',
'description' => 'Remember the famous "more cowbell" skit from Saturday Night Live? Well, this is the actual cowbell.',
'price' => 299.99,
'urls' => array('http://www.nbc.com/Saturday_Night_Live/',
'http://en.wikipedia.org/wiki/More_cowbell')
);
$itemHat = array(
'id' => 'itemHat',
'description' => 'Michael Jackson\'s hat as worn in the "Bille Jean" video. Not really rock memorabilia, but it smells better than Slash\'s tophat.',
'price' => 1699.99,
'urls' => array('http://www.michaeljackson.com/',
'http://music.yahoo.com/vid-2143030--Billie-Jean')
);
$details = array (
'itemGuitar' => $itemGuitar,
'itemShades' => $itemShades,
'itemCowbell' => $itemCowbell,
'itemHat' => $itemHat
);
$itemDetail = $details[$_REQUEST['ImageID']];
$output = $json->encode($itemDetail);
print($output);
?>
500内部服务器错误显示:
Connection close
Content-Encoding gzip
Content-Length 20
Content-Type text/html
Date Sun, 01 Sep 2013 22:47:32 GMT
Server Apache/2.2.22 (Ubuntu)
Vary Accept-Encoding
X-Powered-By PHP/5.3.10-1ubuntu3.7
此代码的问题之一是$.get()
不能按预期方式工作,因为我不断收到500个内部服务器错误。 一旦解决,我就不确定如何将JSON数据提取到包含JSON数据的PHP文件中(请参阅代码中的注释问题)。 有什么解决办法吗?
正如@joshjwalker指出的那样,您可以使用
$itemDetail = $details[$_GET['ImageID']];
echo json_encode($itemDetail);
和你的js脚本可能是
getJSON("getDetailsJSON.php",
{"ImageID" : escape(itemName)},
function(data){
console.log(JSON.stringify(data))
}
);
第一步是找到您的apache错误日志。 这通常会告诉您500服务器错误是什么,因为php端代码中发生了某种错误。
其次,这就是您从php解析json数组的方式,但是您是否正在使用json_encode将php数据编码为php文件中的json?
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