jQuery技术:尝试将使用SQL语句生成的JSON数据从PHP脚本返回到JS网页，但是获取null

我想按下提交按钮，从PHP脚本中的结果SQL语句返回JSON数据，但我收到null 。

我将使用返回的JSON在我的Google地图上过滤显示标记，但是现在我只想将数据从PHP脚本返回到我的jQuery页面，以便我可以操作/使用它。

提交按钮 ：

HTML

JS

 $('#myform').on('submit', function(e) { e.preventDefault(); var myData = $('#myform').serializeArray(); $.getJSON('myscript.php', myData, function(json){ alert(json);// actually filter for later }); }); 

PHP脚本 ：

 // action is a hidden form control I use to check if form was submitted if(isset($_POST["action"])){ if(isset($_POST["color"]) && isset($_POST["zipcode"])){ // try to open a connection to a MySQL server $connection = mysql_connect($host, $username, $password) or die("Could not connect" . mysql_error()); // select the active MySQL database to work with $db_selected = mysql_select_db($database, $connection) or die("Can't use db:" . mysql_error()); $query = 'sql statement to return resutls based on what color and zipcode was provided'; $result = mysql_query($query) or die("Can't do that: " . mysql_error()); } // close connection to the database echo json_encode($result); mysql_close($connection); } 

 ... $result = mysql_query($query); if ( $result === false ) { die("Can't do that: " . mysql_error()); } $retVal = array(); while( $row = mysql_fetch_array( $result ) ) { $retVal[] = $row; } ... echo json_encode( $retVal ); 

 header('Content-Type: application/json'); 

 $.ajax({ url: url, data: myData, success: function(json) {}, error: function(json) {} // this should allow you to check if data is received (but since the content type is set to text/html and $.getJSON expectr application/json it won't be a success) }); 

以上就是jQuery教程分享尝试将使用SQL语句生成的JSON数据从PHP脚本返回到JS网页，但是获取null相关内容,想了解更多jQuery开发(异常处理)及jQuery教程关注计算机技术网(www.ctvol.com)!)。