php - 在PHP中将JSON传递给数组

我正在研究一种可以立即过滤结果的过滤器，我想知道这是否可能是问题的原因，所以我想问一下是否有人可以给我关于如何进行操作的指示。

      <script>
        var services = [
 <?php
//Variables for connecting to your database.
//These variable values come from your hosting account.
$hostname = "###";
$username = "###";
$dbname = "###";

//These variable values need to be changed by you before deploying
$password = "###";
$usertable = "###";
$url = "permalink";
$title = "Address";
$amount = "rent";
$id = "id";
$status = "Beds";
$nonprofit = "Address";

//Connecting to your database
mysql_connect($hostname, $username, $password) OR DIE ("He's dead Jim");
mysql_select_db($dbname);

//Fetching from your database table.
$query = "SELECT * FROM $usertable";
$result = mysql_query($query);

if ($result) {
    while($row = mysql_fetch_array($result)) {
        $url = $row["$url"];
        $title = $row["$title"];
        $amount = $row["$amount"];
        $id = $row["$id"];
        $status = $row["$status"];
        $nonprofit = $row["$nonprofit"];

        echo '{"permalink": "';
        echo "{$url}";
        echo '",';
        echo '"title": "';
        echo "{$title}";
        echo '",';
        echo '"amount":';
        echo "{$amount}";
        echo ',';
        echo '"id": "';
        echo "{$id}";
        echo '",';
        echo '"status": "';
        echo "{$status}";
        echo '",';
        echo '"address": "';
        echo "{$address}";
        echo '",';
        echo '},';

    }
}
?>

]
                //]]>
      </script>


  <script id="template" type="text/html">
       <a title="{{title}}" href="{{permalink}}">
         <div class="fs_box hide-for-small-down">
           <div class="fs_left">
             <span class="fs_head">{{title}}</span>
             <span class="fs_id"><img src="images/{{id}}.jpg" width="75%" height="75%" onError="this.onerror=null;this.src='images/logo.png';"></span>
       <span class="fs_status">{{status}}</span>
             <span class="fs_disc">{{address}}</span>
           </div>
           <div class="fs_price">${{amount}}+</div>
           <div class="clear"></div>
        </div>
      </a>
    </script>

我期望它产生一堆结果，然后将这些结果过滤后的条件显示在页面的其他地方。
当我目前尝试作为php代码时，它的输出很好。但是，当我在php文件中尝试该文件时，它什么也不会产生。还是不喜欢脚本？
谢谢你的帮助！

最佳答案

您可以使用json_decode和json_encode将数组转换为json，然后将json返回数组。

也可能有人会提到您不应该在PHP中使用mysql_ *函数，因为它们已被贬值。

像这样：

 <?php
//Variables for connecting to your database.
//These variable values come from your hosting account.
$hostname = "###";
$username = "###";
$dbname = "###";

//These variable values need to be changed by you before deploying
$password = "###";
$usertable = "###";
$url = "permalink";
$title = "Address";
$amount = "rent";
$id = "id";
$status = "Beds";
$nonprofit = "Address";

//Connecting to your database
mysql_connect($hostname, $username, $password) OR DIE ("He's dead Jim");
mysql_select_db($dbname);

//Fetching from your database table.
$query = "SELECT * FROM $usertable";
$result = mysql_query($query);

if ($result) {
    $results = array()
    while($row = mysql_fetch_array($result)) {
        $results[] = $row;
    }

    $json =  json_encode($results);
}
?>

]

<script>
    var services = <?php echo $json; ?>;
</script>

这将为您提供一个用于在脚本中呈现的json对象。

关于php - 在PHP中将JSON传递给数组，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/26149869/