使用jQuery解析的php编码变量调试不可见的JSON错误

Question

我使用php json_encode向客户端发送一堆值，其中jquery

{"data":{"application":{"basics":{"email":"jepp@mily.com","name_first":"Step","name_last":"Bob","phone":"9210938102938","occupation":"unemployed","website":"wwwcom"},"application":{"title":"Default title","guid":"9as8ua9sd8ua9sdu","language":"sv"},"job":{"title":"joburl","url":"joburl"},"letter":{"letter":"abadbabkbakbakbakbakbakbabk"},"work-experiences":[{"title":"Default WORK Experience Title","url":"wwwsscom","date_from":"1970-01-01","date_to":"1999-12-31"},{"title":"Default WORK Experience Title","url":"wwwsscom","date_from":"1970-01-01","date_to":"1999-12-31"},{"title":"Default WORK Experience Title","url":"wwwsscom","date_from":"1970-01-01","date_to":"1999-12-31"}],"educations":{"title":"Default Education","url":"ixi","date_from":"1970-01-01","date_to":"1999-12-31"},"skills":{"title":"Defailt SKILL","url":"wwwsdcom","date_from":"ixi","date_to":"ixi"},"work-samples":{"title":"A sample of my work","url":"wwwsdcom","date_from":"ixi","date_to":"1999-12-31"}}},"error":[],"warning":[]}

如果我试图在脚本中使用$.parseJSON解析此代码，则不会得到任何对象。但是，如果我将它直接复制/粘贴到控制台中(并添加‘在开头和结尾’)，它就能工作。没有错误代码，我也看不到换行符。JSON lint-工具不返回错误..。

我已经设置了正确的内容类型，并尝试了jquery提供的不同的json解析器。

我错过了什么？

代码是从jQuery教程中剪切/粘贴的。我尝试了一些不同的例子，但都失败了。

var jqxhr = $.getJSON( "application_controller.php", function() {
    console.log( "success" );
})
.done(function() {
    console.log( "second success" );
})
.fail(function() {
    console.log( "error" );
})
.always(function(data) {
    console.log( "complete" );
    application = data;  
});

// Perform other work here ...

// Set another completion function for the request above
jqxhr.complete(function() {
    console.log( "second complete" );
});
});

是真的，我在控制台里破译的。这是一个在脚本中执行的片段(它也不能工作)：

$.ajax({
  dataType: "json",
  contentType: "application/json",
  url: 'application_controller.php',
  data: '{id:id}',
  success: function( data ) {
        application = data.responseText;
        application = $.parseJSON(application);/* < string */     
    },
    fail: console.log("fail"),
    complete: function(data) {
        console.log(data.responseText);
        application = data.responseText;
    }
});

Answer 1

回答我自己的问题：

为了解决这个错误，我不得不将我所有的php文件从"utf8“编码更改为”没有BOM的utf8“。然后起作用了。

当包含一个没有utf8编码的文件(当然，在层次结构中很低)时，它会污染所有其他文件并破坏输出。

Answer 2

使用JSON.parse(jsonString)函数。

var jsonString = '{"data":{"application":{"basics":{"email":"jepp@mily.com","name_first":"Step","name_last":"Bob","phone":"9210938102938","occupation":"unemployed","website":"wwwcom"},"application":{"title":"Default title","guid":"9as8ua9sd8ua9sdu","language":"sv"},"job":{"title":"joburl","url":"joburl"},"letter":{"letter":"abadbabkbakbakbakbakbakbabk"},"work-experiences":[{"title":"Default WORK Experience Title","url":"wwwsscom","date_from":"1970-01-01","date_to":"1999-12-31"},{"title":"Default WORK Experience Title","url":"wwwsscom","date_from":"1970-01-01","date_to":"1999-12-31"},{"title":"Default WORK Experience Title","url":"wwwsscom","date_from":"1970-01-01","date_to":"1999-12-31"}],"educations":{"title":"Default Education","url":"ixi","date_from":"1970-01-01","date_to":"1999-12-31"},"skills":{"title":"Defailt SKILL","url":"wwwsdcom","date_from":"ixi","date_to":"ixi"},"work-samples":{"title":"A sample of my work","url":"wwwsdcom","date_from":"ixi","date_to":"1999-12-31"}}},"error":[],"warning":[]}';

var myData = JSON.parse(jsonString);

小提琴