jQuery VM603:1 未捕获的语法错误：JSON 中的意外标记 o 位于位置 1？
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VM603:1 Uncaught SyntaxError: Unexpected token o in JSON at position 1?
提问by Joe.Herylee
n.parseJSON @ jquery-1.12.3.min.js:4
showTable @ query.html:107
success @ query.html:98
i @ jquery-1.12.3.min.js:2
fireWith @ jquery-1.12.3.min.js:2
y @ jquery-1.12.3.min.js:4
c @ jquery-1.12.3.min.js:4
Above is the error info of code.
以上是代码的错误信息。
I want to parse JSON object to array,and render the array into the HTML. But I don't know how to handle it by the following?
我想将 JSON 对象解析为数组，并将数组呈现到 HTML 中。但我不知道如何通过以下方式处理它？
Following is the part js code of mine , function of showTable(data) is parse the json object into html table.function requestData() is request info from the back.
下面是我的部分js代码，showTable(data)函数是将json对象解析成html table。函数requestData()是后面的请求信息。
(function(){
var form = $("form");
var contentCenter = $(".content-center");
$(".s-btn").on("click",function(event){
$(".container").css({
"position":"relative"
})
form.css({
"position":"absolute",
"left":"15px",
"top":"0px"
});
contentCenter.css({
"position":"absolute",
"top":"-12px"
})
event.preventDefault();
requestData();
});
//与后台交互数据
function requestData(){
var data = {
type : $("#form_control").val(),
keywords : $.trim($("#ipt_control").val())
};
$.ajax({
type : "GET",
url : "data/data.json",
dataType : "json",
data : data,
success:function(msg){
//TODO请求成功后跳转页面
//处理后台返回的数据
console.log(msg);
showTable(msg);
},
error:function(msg){
console.log("failed");
}
});
}
//获取json数据，动态创建表格
function showTable(data){
var dataArray = $.parseJSON(data);
console.log(dataArray);
var tableStr="<table class='table table-bordered'>"
tableStr = tableStr + "<thead><td>id</td><td>name</td>handle<td></td>"
var len = dataArray.length;
for(var i=0;i<len;i++){
tableStr = tableStr + "<tr><td>" + dataArray[i].id + "</td>" + "<td>" + dataArray[i].name + "</td>" + "<td>" + dataArray[i].handle + "</td></tr>";
}
tableStr = tableStr + "</table>"
$("#dataType").html(tableStr);
}
})();
回答by Danielle Davis
I had this same problem, if your response header is application/json, it is already parsed, you don't need to parse it with:
我有同样的问题，如果你的响应头是 application/json，它已经被解析，你不需要解析它：
var dataArray = $.parseJSON(data);
Also, you can get it with jQuery.getJSON(): jQuery.getJSON()
此外，您可以通过以下方式获取它jQuery.getJSON()：jQuery.getJSON()
回答by Stéphane Molano
In your ajax request you should specify the dataType as "html". Among others : when data is generated by php json_encode() function.
在您的 ajax 请求中，您应该将 dataType 指定为“html”。其中：当数据由 php json_encode() 函数生成时。
$.ajax({
type : "GET",
url : "data/data.json",
dataType : "html",
data : data,
success:function(msg){
console.log(msg);
var array_return = $.parseJSON ( msg );
},
error:function(msg){
console.log("failed");
}
});
回答by Waheed
So here is the trick with JSON parser
所以这是使用 JSON 解析器的技巧
JSON.parse function which is mostly used in $.parseJSON(data) from jquery. Will always assume a string starting with ' a single quote.
JSON.parse 函数，主要用于 jquery 的 $.parseJSON(data) 中。将始终假定以 ' 开头的字符串是单引号。
So if your data looks like this
所以如果你的数据看起来像这样
let ok = '{"hello": "world"}'; //success
let fail = "{'hello': 'world'}"; //fail
and then
进而
JSON.parse(ok);
JSON.parse(fail);
With a single quote it will work just as expected, otherwise it will fail with Unexpected token o in JSON at position 1?and the reason is Single quotes (') are not allowed in JSON as safari explains.
使用单引号它将按预期工作，否则它将失败，Unexpected token o in JSON at position 1?原因是 safari 解释说，JSON 中不允许使用单引号 (')。
Hope that helps :)
希望有帮助:)
