Question

我想知道如何在返回时为每个JSON项添加一个html链接。

php文件

//relevant code, $result contains sql query with records from a search
    $records = array();
    while($array = mysqli_fetch_array($result))
     {
        $records[] = $array;
     }
   echo(json_encode($records));

通过ajax输出html

function responseReceived(e){
        document.getElementById('response').innerHTML = e.target.responseText;
    }

响应是div标记。所以我想知道如何将html链接（到另一个页面）添加到json中的每个项目。因为目前它输出json，但它不允许我添加html。这是当前输出，由于我通过表单输入数据，这是预期的，我只想在他旁边添加一个小html链接。

[{"0":"1115","ID":"1115","1":"john","name":"john","2":"male","type":"male","3":"berk","country":"berk","4":"aus","region":"aus","5":"32.43","lon":"32.43","6":"32.54","lat":"32.54","7":"nothing to say","description":"nothing to say"}]

提前谢谢。

Answer 1

你从哪里获得链接？

返回的数据是链接还是返回的数据应该格式化为链接的文本？

如果是第二个，你需要从某个地方获取链接。

你可以做的是在数据库中存储某种链接/ id并在json中打印出来，这样就会输出如下内容：

[{"url": "http://example.com", "content": "This is a link!"}]

我认为你知道如何进行ajax调用，所以让我们继续格式化：

// Let's make a function
function createLinks(json){

    // Let's parse the JSON first (if it's a string)
    json = JSON.parse(json);

    // Loop through all the elements
    for(var i = 0; i < json.length; i++){

        // Check if the fields exist
        if(json[i].url && json[i].content){

            // Creating a tag
            var a = document.createElement("a");

            // Let's add the values to the a tag
            a.href = json[i].url;
            a.textContent = json[i].content;
            // ^ or innerHTML if you want to have HTML code there

            // Appending to body element (just for this example)
            document.body.appendChild(a);

        }

    }

}

我希望它有所帮助！

Answer 2

这个简单的代码添加了每个数组项的链接。如果要将其转换为HTML，只需创建HTML字符串，然后可以使用innerHTML插入它。

var obj = JSON.parse('[{"0":"1115","ID":"1115","1":"john","name":"john","2":"male","type":"male","3":"berk","country":"berk","4":"aus","region":"aus","5":"32.43","lon":"32.43","6":"32.54","lat":"32.54","7":"nothing to say","description":"nothing to say"},{"0":"1115","ID":"1115","1":"john","name":"john","2":"male","type":"male","3":"berk","country":"berk","4":"aus","region":"aus","5":"32.43","lon":"32.43","6":"32.54","lat":"32.54","7":"nothing to say","description":"ggg"}]');
for (var i = 0; i < obj.length; i++) {
    obj[i].link = '<a href="http://www.google.com">Google</a>';
}

如何通过php和ajax将html添加到返回的json对象中

2 个答案: