javascript 语法错误：JSON 解析错误：意外的标识符“对象”
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SyntaxError: JSON Parse error: Unexpected identifier "object"
提问by Vasvas
Please help me to understand what's wrong. I want to parse JSON reply as object.
请帮助我了解出了什么问题。我想将 JSON 回复解析为对象。
PHP process.php code:
PHP process.php 代码：
<?php
$return = array();
array_push($return['amount']="$amount");
array_push($return['fee']="$fee");
array_push($return['total']="$total");
echo json_encode($return);
?>
Returns JSON string:
返回 JSON 字符串：
{"amount":"350","fee":"0","total":"350"}
JS (jquery) code:
JS（jquery）代码：
$.getJSON("process.php?amount="+amount, function(data,status) {
var obj = $.parseJSON(data);
alert (obj.amount);
});
I receive error:
我收到错误：
SyntaxError: JSON Parse error: Unexpected identifier "object"
语法错误：JSON 解析错误：意外的标识符“对象”
BUT! When I try to insert result instead data (but insert ' quotes left/right):
但！当我尝试插入结果而不是数据时（但插入 ' 左/右引号）：
var obj = $.parseJSON('{"amount":"350","fee":"0","total":"350"}');
And I see alert = 350. So, it's working good.
我看到 alert = 350。所以，它运行良好。
I try to make something like that:
我试着做这样的事情：
var jsonreply = "'"+data+"'";
var obj = $.parseJSON(jsonreply);
But received below error:
但收到以下错误：
SyntaxError: JSON Parse error: Single quotes (') are not allowed in JSON
语法错误：JSON 解析错误：JSON 中不允许使用单引号 (')
回答by icktoofay
getJSONparses the JSON for you —?calling $.parseJSONwill convert the object into the string [object Object]and then try to parse that, giving you an error. Just omit the $.parseJSONcall and use datadirectly.
getJSON为您解析 JSON —?calling$.parseJSON会将对象转换为字符串[object Object]，然后尝试解析它，给您一个错误。直接省略$.parseJSON调用data即可。
Furthermore, I should note that the calls to array_pushare strange and unnecessary. array_pushusually takes an array and a value to push on to it, but (for example) in your first line you're setting $return['amount']to "$amount"and then passing $return['amount']to array_push, which does nothing at best and might give you a warning or error at worst. You'd get the exact same behavior if you did this:
此外，我应该注意到对 的调用array_push很奇怪且不必要。array_push通常需要一个数组和一个值来推送它，但是（例如）在你的第一行中你设置$return['amount']为"$amount"然后传递$return['amount']给array_push，这充其量什么都不做，最坏的情况可能会给你一个警告或错误。如果你这样做，你会得到完全相同的行为：
$return['amount']="$amount";
$return['fee']="$fee";
$return['total']="$total";
Then you might also realize that the quotes around, say, "$amount"are unnecessary, and you could actually do this:
然后你可能也会意识到周围的引号"$amount"是不必要的，你实际上可以这样做：
$return['amount']=$amount;
$return['fee']=$fee;
$return['total']=$total;
Finally, you can actually condense all five lines using some special arraysyntax very easily:
最后，您实际上可以使用一些特殊array语法非常轻松地压缩所有五行：
echo json_encode(array(
'amount' => $amount,
'fee' => $fee,
'total' => $total
));
This is quite a bit nicer if I do say so myself.
如果我自己这么说，那就更好了。
回答by Vicky Gonsalves
Actually u dont need to parse it. U can directly access it
其实你不需要解析它。你可以直接访问它
$.getJSON("process.php?amount="+amount, function(data,status) {
alert (data.amount);
});
回答by lolrs2013
It seems like your error is here:
您的错误似乎在这里：
var jsonreply = "'"+data+"'";
Try to escape those ' with "\". Like
尝试用“\”转义那些 '。喜欢
var jsonreply = "\'"+data+"\'";
