将JSON从PHP传递到JavaScript变量
[英]Passing JSON from PHP to a JavaScript variable
问题描述
I'm making an aplication with phonegap and I'm stuck trying to send JSON data from the PHP on the server to JavaScript on the device. 
我正在使用phonegap进行复制，并且尝试将JSON数据从服务器上的PHP发送到设备上的JavaScript时遇到了麻烦。 I want to do something like: 我想做类似的事情：
var JSON = '{ "Table" : ' + "http://www.hel.net/LoadDB.php=?table=exhibitions" + '}';
the php works fine and returns somethig like: PHP工作正常，并返回somethig像：
"[{"id":"1","name":"Art for everyone","image":null,"name2":"June 29, 2013: 11:00am.","description":"With reservation\r\nFree entrance","description2":null}]"
I want that result in a javascript variable to work later with: 我希望该结果可在javascript变量中稍后使用：
var obj = eval ("(" + JSON + ")");
document.getElementById("rName").innerHTML=obj.Table[1].name;
document.getElementById("lname").innerHTML=obj.Table[1].name2;
What I want to do is something like: 我想做的是这样的：
var JSON = '{ "Table" : ' + "http://www.hel.net/LoadDB.php=?table=exhibitions" + '}';
var obj = eval ("(" + JSON + ")");
document.getElementById("rName").innerHTML=obj.Table[1].name;
document.getElementById("lname").innerHTML=obj.Table[1].name2;
How can I make the first line work? 我如何使第一线工作？ is it possible to make the first line work? 是否可以使第一线工作？ PS. PS。 I do not have much experience with JSON arrays. 我对JSON数组没有太多经验。
Ok I tried ajax and works, I used: 好的，我尝试了ajax并工作了，我使用了：
console.log("before");
var jqxhr = $.ajax( "http://www.hel.com/LoadDB.php?table=exhibitions" )
.done(function(data) { console.log(data); })
.fail(function() { console.log("error"); })
.always(function() { console.log("complete"); });
console.log("after");
more info in: 更多信息：
3 个解决方案
解决方案1
1 2013-06-27 01:47:14
I think all you need is var obj = <?php echo $myjsonencodedvar; ?> 我认为您只需要var obj = <?php echo $myjsonencodedvar; ?> var obj = <?php echo $myjsonencodedvar; ?>
or 要么
var obj = <?php echo json_encode($myarray_or_object); ?>
Since I said "I think..." I decided to test it out. 自从我说“我认为...”以来，我决定进行测试。 I found the following dump() function here on SO. 我在SO上找到了以下dump（）函数。
$arr=array(1,'biscuit'=>'gravy',3,4,5);
$json=json_encode($arr);
?>
<script>
j=<?php echo $json; ?>;
document.write(dump(j));
function dump(obj) {
var out = '';
for (var i in obj) {
out += i + ": " + obj[i] + "\n";
}
return out;
}
</script>
output: 输出：
0: 1 biscuit: gravy 1: 3 2: 4 3: 5
解决方案2
0 2013-06-27 01:49:59
使用JSONP（无回调），并在客户端使用$ .getJSON（），它将从json字符串解析为js对象。
解决方案3
0 已采纳 2013-06-27 02:34:42
Try this: 尝试这个：
PHP: (json.php) PHP：（json.php）
<?php
header("Content-Type: text/json");
//the data format your question mentioned
$data = array("Table"=>array(array("id"=>"1","name"=>"Art for everyone","image"=>null,"name2"=>"June 29, 2013","description"=>"With reservation\r\nFree entrance","description2"=>null)));
echo json_encode($data);
?>
Front-end: 前端：
<!DOCTYPE html>
<html>
<head>
<title></title>
</head>
<body>
<script src="//ajax.googleapis.com/ajax/libs/jquery/1.10.1/jquery.min.js"></script>
<script>
$.get("json.php",function(json){
alert(json.Table[0].name);
});
</script>
</body>
</html>
Hope this is helpful for you. 希望这对您有帮助。
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