Question

我试图在PHP中选择选项值，我想比较选定的值和数据库值，例如PHP中的seat id。但是，它会导致未定义的错误。你能查一下我的代码吗？

<script type="text/javascript">
function transfer(){ 
 var pix = $("#pix option:selected").val(); 
    document.abc.test.value =pix; 
  } 
</script>
    <script                  
  src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js">      
 </script>
 <script>
      function ajaxWay() {
        var sel = $("#pix option:selected").val();
   $.get("p.php?pix="+ sel );
     // syntax: $.post(URL,data,callback);
       $.get("p.php",  function(dataFromtheServer) {
       $("#result").html(dataFromtheServer);
     });             
    }
  </script>
   <body>
  <div id="a" style="text-align:center;">
      <?php
  $servername = "";
  $username = "";
  $password = "";
  $dbname = "";

  // Create connection
  $conn = mysqli_connect($servername, $username, $password, $dbname);
  // Check connection
  if (!$conn) {
      die("Connection failed: " . mysqli_connect_error());
  }   

  $sql = "SELECT * FROM seat" ;            

  $result = mysqli_query($conn, $sql) or die("Query: ($sql) [problem]");  
  ?>
   <form name="abc"  method="get">
    <select name="select" id="pix">
        <?php

            if (mysqli_num_rows($result) > 0) 
             {   
            while($row = mysqli_fetch_assoc($result)) 
              {                                     
                echo "<option value='".$row[seat_id]. "'>" .$row[seatNumber].  "</option>";
              }
          } 
           mysqli_close($conn);    
        ?>
    </select>
    <input type="button" value="click" onclick="transfer();">    
     <input type="button" value="click" onclick="ajaxWay()">    
    <input type="text" name="test" id="text"> 
    <aside id="result">i will put the empty space from database</aside>
</form>

p.php

<?php
$servername = "***.compute-1.amazonaws.com";
$username = "***";
$password = "***";
$dbname = "***";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
    die("Connection failed: " . mysqli_connect_error());
}

$sql = "SELECT * FROM seat";
$result = mysqli_query($conn, $sql) or die("Query: ($sql) [problem]");
$fields = mysqli_num_fields($result);

if (mysqli_num_rows($result) > 0) {
    // output data of each row
   // display("<table border=1><th>firstname</th><th>lastname</th><th>items</th>","\n");

    while($row = mysqli_fetch_assoc($result)) {

        if($_GET['pix'] == $row['seat_id']){

        if($row['seatStatus'] == 1) 
                {
                    echo "available" ;
                }       
                else
                {

                    echo "unavailable";
                }    
                }
    }

  } else {
       echo "0 results";
  }
  mysqli_close($conn);    

?>

Answer 1

这是一个猜测，但它看起来有2个GET请求，参数和另一个没有 - 回调函数似乎正在使用没有参数的调用所以它可能看起来更多像：

$.get("p.php?pix="+ sel,  function( dataFromtheServer ) {
    $("#result").html( dataFromtheServer );
});

我不知道以下内容是否会有所帮助，但我不禁觉得ajax功能看起来很奇怪。

<html>
    <head>
        <title></title>

        <script type="text/javascript">
            function transfer(){ 
                var pix = $("#pix option:selected").val(); 
                document.abc.test.value=pix; 
            } 
        </script>
        <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.2/jquery.min.js"></script>

        <script>
            function ajaxWay() {
                var sel = $("#pix option:selected").val();
                console.info( 'send ajax request: p.php?pix=%s', sel );

                $.get("p.php?pix="+ sel,  function(r) {
                    console.info('response: %s',r );
                    $("#result").html( r );
                });             
            }
        </script>
    </head>
    <body>
        <div id="a" style="text-align:center;">
        <?php

            $servername = "";
            $username = "";
            $password = "";
            $dbname = "";

            $conn=new mysqli($servername, $username, $password, $dbname);
            $sql='select * from `seat`;';
            $result = $conn->query( $sql );  
        ?>
        <form name="abc" method="get">
            <select name="select" id="pix">
                <?php                                   
                    if( $result ){
                        while( $rs=$result->fetch_object() ){
                            echo "<option value='{$rs->seat_id}'>" . $rs->seatNumber;
                        }
                    }
                    $conn->close();    
                ?>
            </select>
            <input type="button" value="click" onclick="transfer();">    
            <input type="button" value="click" onclick="ajaxWay()">    
            <input type="text" name="test" id="text"> 
            <aside id="result">i will put the empty space from database</aside>
        </form>
    </body>
</html>


/* p.php */
$servername = "ec2-52-201-212-193.compute-1.amazonaws.com";
$username = "s2016b_user21";
$password = "s2016b_user21";
$dbname = "s2016b_user21";

/* alternative sql */
$conn = new mysqli( $servername, $username, $password, $dbname );
$sql='select `seatStatus` from `seat` where `seat_id`=?;';
$stmt=$conn->prepare( $sql );
$stmt->bind_param( 'i', $_GET['pix'] );

if( $stmt ){

    $stmt->execute();

    $stmt->bind_result( $status );
    $stmt->fetch();

    echo $status==1 ? 'available' : 'unavailable';
}

$conn->close();

Answer 2

我不太了解JQuery 您可以尝试在javascript使用console.log(sel)，在php使用var_dump($_GET['pix'])以确保已获取“sel”的值。

//javascript
var pix = $("#pix option:selected").val();
console.log(pix);
//PHP
var_dump($_GET['pix']);

如果var_dump($_GET['pix'])的结果为NULL，那么您的javascript代码看起来应该更像：

 <script>
  function ajaxWay() {
    var sel = $("#pix option:selected").val();//check the value of sel
    var url = "p.php";//check the path of p.php
    $.get(url,{'pix':sel},function(dataFromtheServer) {
       $("#result").html(dataFromtheServer);
    });             
  }
</script>

如何获得选定的选项值？

2 个答案: