Question

我正在尝试将第二个数据源中的选择插入到“自动完成”下拉列表中。第一和第二源提取都是通过ajax。我精心编写了一个JSON数组，用于将2d源数据插入到Autocomplete的下拉列表中。它通过了jsonLint测试。格式如下：

[{"label":"New York, Bronx, Bronx County, New York, United States","value":"Bronx, Bronx County, New York, United States"},
{"label":"New York, Brooklyn, Kings County, New York, United States","value":"Brooklyn, Kings County, New York, United States", }]

我在创建代码以将变量数据放入该类型和格式时遇到了很大的麻烦。数据库当前包含JSON字符串。我使用PHP的json_encode（）来创建传递给浏览器的数据对象。数据库条目可以更改，但现在看起来像：

"label":"New York, Bronx, Bronx County, New York, United States","value":"Bronx, Bronx County, New York, United States"

似乎1. JSON字符串不会更改为对象，并且2.某些外来的未知字符会以某种方式插入。但是，我是一个新手，特别是功能和对象。这是jQuery / Javascript，以及关于各点数据状态的一些评论。

var boroughData = function() {
    $.ajax({
        //normal stuff like url:, dataType:, etc. It works.
        success: function (data){
            boroughData = $.map( data, function (item){
                //data is JSON, but info inside extra " ", like Object { boroughString=""label":"Dallas, Cockre..., Texas, United States""}
                // # response like [{"boroughString":"\"label\":\"Dallas, Cockrell Hill, Dallas County, Texas, United States\", . . ."}]
                return item.boroughString;//returns JSON strings, not objects, in an array
            });
            alert(jQuery.isArray(boroughData)  + "|bD1"); //true
            console.log(boroughData); //array containing JSON strings in red
        } //closes success: for boroughData
    }); //closes ajax for boroughData
}(); //closes boroughData and (); executes it
alert(jQuery.isArray(boroughData)  + "|bD2"); //false
alert(boroughData + "|bD3"); //"label":"Dallas, Cockrell Hill, Texas", "value":"Dallas, Cockrell Hill, Texas","label":"Dallas, Downtown Dallas, Texas", "value":"Dallas, etc.
alert(JSON.parse(boroughData) + "|bD4");//SyntaxError: JSON.parse: unexpected non-whitespace character after JSON data

请注意，bD1处的isArray测试为true，而bD2处的isArray测试为false，它们之间唯一发生的事情是关闭函数和ajax。另请注意带有散列＃标记的注释行，用于记录看似额外的“”标记。

如何让boroughData函数输出一个JSON数组，其格式与我在上面第一个代码框中显示的格式相同？请具体说明您的建议和代码。我是新手，不遵循一般说明。

Answer 1

IMO您的函数不返回值。我不确定你的方式是否真的能够正常执行，但请相信：）

    function boroughData () {
    $.ajax({
        success: function (data){
            boroughData = $.map( data, function (item){
                return item.boroughString;//returns JSON strings, not objects, in an array
            });
            alert(jQuery.isArray(boroughData)  + "|bD1"); //true
            console.log(boroughData); 
        } 
    });
    return boroughData; // it's not good practice to use the same name for local variable... 
    }(); 
    alert(jQuery.isArray(boroughData)  + "|bD2"); //false
    alert(boroughData + "|bD3");
    alert(JSON.parse(boroughData) + "|bD4");

jQuery .map（）函数的类型和格式问题

1 个答案: