Question

这是我的功能

function getRecords(){            
    $.getJSON( "viewcustomers.php", function(data) {
        var items = [];
        $.each( data, function( key, val ) {
            items.push( "<td>" + val + "</td>" );
            alert(items);
         });
    });
}

viewcustomers.php中的数据如下：

[{"id":"1","name":"ali","cnic":"01","address":"nipa","email":"301","phone":"luxairy "},{"id":"2","name":"raheel","cnic":"1234567","address":"abc flat","email":"raheel@gmail.com","phone":"0342"},{"id":"5","name":"Raheel","cnic":"123-4567-45","address":"Gulshan block 13-c","email":"raheel@gmail.com","phone":"03422301775"},{"id":"4","name":"zain","cnic":"12345678","address":"f.b area","email":"zain@gmail.com","phone":"7894561"},{"id":"6","name":"Raheel","cnic":"123-4567-45","address":"Gulshan block 13-c","email":"raheel@gmail.com","phone":"03422301775"},{"id":"7","name":"Ali","cnic":"123456786","address":"nipa","email":"ali@gmail.com","phone":"1234568"},{"id":"8","name":"Waqas","cnic":"123456748","address":"F.b area","email":"waqas@gmail.com","phone":"1234568"},{"id":"9","name":"Owais","cnic":"1234567","address":"Fb area","email":"owais@gmail.com","phone":"12345678"}]

但是当我运行我的jquery函数时，它给了我[object Object]而不是上面的键值。

Answer 1

在您的情况下，item实际上是array。而你正在从JSON中推动其中的值。所以为了以字符串格式查看它，

你必须使用，

alert(JSON.stringify(items));

在深入了解您的JSON时，您正在遍历每个对象，因此您应该在插入数组时对这些对象进行字符串化，例如，

items.push( "<td>" + JSON.stringify(val) + "</td>" );

Answer 2

因为对象的to字符串实现返回[object Object]，所以您可以使用JSON.stringify()来处理此问题。对于旧浏览器，您必须包含json2

之类的polyfil

items.push( "<td>" + JSON.stringify(val) + "</td>" );

演示：Fiddle - 查看浏览器控制台

<强>更新

如果您使用.map()

，应该很简单

var items = $.map(data, function (obj) {
    return "<tr>" + $.map(obj, function (val) {
        return '<td>' + val + '</td>'
    }).join('') + "</tr>";
});
console.log(items.join(''));

演示：Fiddle

返回[object，Object]而不是正确的JSON

2 个答案: