Question

我有一个AJAX请求来显示用户邮政编码的地址列表。它将在下拉列表中显示它们供用户选择。我在下面展示了PHP，JS和repsonse。

我使用cURL获得了JSON响应。我正在使用CI和jQuery。

如何从响应中提取我需要的数据并将它们放入HTML选项标记中。 ID将出现在选项值标签中，描述将介于开始和结束选项标签之间。

这是我的JS：

$.get(
    '<?=site_url('api/postcode') ?>',
    {
        'postcode' : postcode
    },
    function(data) {

        $( '#loader-address' ).hide();

        // Show the dropdown list           
        var list = '<option value="">Choose...</option>';

        // Create address options from JSON

    },
    'json'
);

这是我的PHP函数：

public function get_by_postcode( $postcode )
{
    /* Build up the URL to send the request to. */
    $sURL = "http://services.Postcodeanywhere.co.uk/json.aspx?";
    $sURL .= "account_code=" . urlencode($this->pca->account);
    $sURL .= "&license_code=" . urlencode($this->pca->license);
    $sURL .= "&action=lookup";
    $sURL .= "&type=by_Postcode";
    $sURL .= "&postcode=" . urlencode($postcode);


    $ch = curl_init();
    $timeout = 5;
    curl_setopt($ch,CURLOPT_URL,$sURL);
    curl_setopt($ch,CURLOPT_RETURNTRANSFER,1);
    curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,$timeout);
    $ContentsFetch = curl_exec($ch);
    curl_close($ch);
    return $ContentsFetch;
}

这是JSON respone：

"[{\"id\":\"5378660.00\",\"seq\":\"0\",\"description\":\"1 Carrington Close Croydon\"},\r\n{\"id\":\"5378661.00\",\"seq\":\"1\",\"description\":\"2 Carrington Close Croydon\"},\r\n{\"id\":\"5378662.00\",\"seq\":\"2\",\"description\":\"3 Carrington Close Croydon\"},\r\n{\"id\":\"5378663.00\",\"seq\":\"3\",\"description\":\"4 Carrington Close Croydon\"},\r\n{\"id\":\"5378664.00\",\"seq\":\"4\",\"description\":\"5 Carrington Close Croydon\"},\r\n{\"id\":\"5378665.00\",\"seq\":\"5\",\"description\":\"6 Carrington Close Croydon\"},\r\n{\"id\":\"5378666.00\",\"seq\":\"6\",\"description\":\"7 Carrington Close Croydon\"},\r\n{\"id\":\"5378667.00\",\"seq\":\"7\",\"description\":\"8 Carrington Close Croydon\"}]\r\n"

Answer 1

您可以在javascript中使用data.id。结合jQuery $.each()函数迭代数据。像这样：

$.get(
    '<?=site_url('api/postcode') ?>',
    {
        'postcode' : postcode
    },
    function(data) {

        $( '#loader-address' ).hide();

        // Show the dropdown list           
        var list = '<option value="">Choose...</option>';
        $.each(data, function(key, value) {
            list += '<option value="'+value.id+'">'+value.description+'</option>';
        });

        // Create address options from JSON

    },
    'json'
);

Answer 2

使用getJSON http://api.jquery.com/jQuery.getJSON/

$.getJSON(url, param ,function(data) { //url - return json page

  $( '#loader-address' ).hide();

   var list = '<option value="">Choose...</option>';
   $.each(data, function() {
       list += '<option value="'+this.id+'">'+this.description+'</option>';
   });
};

javascript中的json响应必须为"[{"id":"5378660.00","seq":"0","description":....。逃脱报价不需要

如何从JSON响应中提取数据

2 个答案: