Question

我正在尝试在JSON对象数组中转换我的邮政编码和城镇，但我想我做得不对，我需要它来实现自动完成功能。

这是我的代码：

var request_image = function(url) {
    return new Promise(function(resolve, reject) {
        var img = new Image();
        img.onload = function() { resolve(console.log("sdfsdf")); };
        img.onerror = function() { reject(console.log("wwwwwwww")); };
        var imageSource = url + '?random-no-cache=' + Math.floor((1 + Math.random()) * 0x10000).toString(16);
        $(img).on("load",loadHandler).on("error",errorHandler)[0].src = imageSource;
        console.log(img.src)
    });
};


var ping = function(url, multiplier) {
    return new Promise(function(resolve, reject) {
        var start = (new Date()).getTime();

        var response = function() { 
            var delta = ((new Date()).getTime() - start);
            delta *= (multiplier || 1);
            resolve(delta); 
        };

        var errorResponse = function() {
            var delta = "Could not connect"
            reject(delta);
        };
        request_image(url).then(response).catch(errorResponse);

        // Set a timeout for max-pings, 5s.
        setTimeout(function() { reject(Error('Timeout')); }, 5000);
    });
};

var errorHandler = function(event) {
    sleep(3000);
    console.log("errorHandler");
    console.log(event);
};

var loadHandler = function(event) {
    sleep(3000);
    console.log("loadHandler");
    this.appendTo("#imgHolder");

};

var sleep = function(milliseconds) {
    var start = new Date().getTime();
    while(1) {
        if ((new Date().getTime() - start) > milliseconds) {
            break;
        }
    }
};

Answer 1

您需要将对象中的每个条目都设为一个数组。这应该有效：

while($row=mysqli_fetch_array($result)){
     $matches[] = array(
                'label'=> $row["postcode"],
                'value'=> $row["town"],
                    );
}

Answer 2

请勿尝试使用字符串插入json。你可以完全依赖json_encode。

我会这样做

 $sql = "SELECT * FROM uk_postcodes";
 $result = mysqli_query($connection, $sql) or die("Error " . mysqli_error($connection));

 $dname_list = array();
 while($row = mysqli_fetch_array($result))
 {
    $dname_list[] = array(
       "label" => $row['postcode'],
       "value" => $row['town']
    );
 }
 header('Content-Type: application/json');
 echo json_encode($dname_list);

转换对象数组

2 个答案: