如何从PHP正确调用jquery函数？

Question

更新：

我正在制作一个喜欢/不喜欢的投票程序，它在Firefox 3，Chrome和Safari中按照我想要的方式工作。但是，它似乎在Firefox 4中不起作用。例如，我创建了一个表单，在用户单击提交按钮并将数据发送到SQL表之后，应该播放jquery动画。这是完整的代码：

<script type="text/javascript" src="opacity.js"></script>



<link href="voter.css" rel="stylesheet" type="text/css" />

<?php


mysql_connect("","","") or die("Error connecting to db");
mysql_select_db("") or die("Error connecting to db");

$currAdd2 = $_SERVER["REQUEST_URI"];
$currAdd2 = substr($currAdd2, 0, -9);
$currAdd2 = substr($currAdd2, 11);

$numrow1 = mysql_query("SELECT * FROM rate WHERE apt = '$currAdd2' AND likes = 1 ");
$numlikes = mysql_num_rows($numrow1); 

$numrow2 = mysql_query("SELECT * FROM rate WHERE apt = '$currAdd2' AND likes = 0 ");
$numdislikes = mysql_num_rows($numrow2);

$numrow3 = mysql_query("SELECT * FROM rate WHERE apt = '$currAdd2'  ");
$totalvotes = mysql_num_rows($numrow3);

$likewidth = ($numlikes/($totalvotes+1)) * 400;
$dislikewidth = ($numdislikes/($totalvotes+1)) * 400;
?>
<form action= '<?php echo $_SERVER['PHP_SELF']; ?>'; method='post'>
<table align='left' width='500px' border='0' cellpadding='0' cellspacing='0'>
<? echo"      
      <tr>
        <td>
            <div id='blue'><input type='image' name='blue' id='likebutton' value='blue' src='../zrating/like.png' 

            /></div>

        </td>
        <td>
        <b>$numlikes Like</b>
        <div id='like' style='width:$likewidth'><div id='dis'></div></div><br>
        </td>
      </tr>
      <tr>
        <td>
            <div id='red'><input type='image' name='red' id='dislikebutton' value='red' src='../zrating/dislike.png'  

            /></div>

        </td>
        <td>
            <b>$numdislikes Dislike</b>
            <div id='dislike' style='width:$dislikewidth'><div id='dis'></div></div><br>
        </td>
      </tr>
      <tr>
      <td>
      </td>
      <td>
      <div id='success'><br><br>Thanks for voting!</div>
      <div id='fail'><br><br>You are only allowed 1 vote per apartment.</div>     
      </td>
      </tr>
  </table>
</form>
";


//check ip address
    if (!empty($_SERVER['HTTP_CLIENT_IP']))   //check ip from share internet
    {
      $ip=$_SERVER['HTTP_CLIENT_IP'];
    }
    elseif (!empty($_SERVER['HTTP_X_FORWARDED_FOR']))   //to check ip is pass from proxy
    {
      $ip=$_SERVER['HTTP_X_FORWARDED_FOR'];
    }
    else
    {
      $ip=$_SERVER['REMOTE_ADDR'];
    }



if ( isset( $_POST['blue'] ) )
{

    $currAdd = $_SERVER["REQUEST_URI"];
    $currAdd = substr($currAdd, 0, -9);
    $currAdd = substr($currAdd, 11);

    $ipquery = mysql_query ("SELECT * FROM rate WHERE ip = '$ip' AND apt = '$currAdd' ");
    $ipquery = mysql_fetch_assoc($ipquery);
    $ipquery = $ipquery['ip'];


    if ($ipquery == $ip)
    {
        echo "<script type='text/javascript' src='jquery.js'></script>";


                echo "<script type='text/javascript'>
                 $(document).ready(function()
    {
    $('#fail').animate({width:300},1000);
    $('#fail').animate({height:100},1000);
    $('#fail').animate({width:300,height:100},1400);
    $('#fail').fadeOut(5000);

    }); </script>";

    }
    else
    {

    mysql_query ("INSERT INTO rate VALUES ('','$currAdd', '1', '$ip')");            

    $rankqry = mysql_query ("SELECT * FROM rank WHERE apt = '$currAdd' ");
    $rankqry = mysql_fetch_assoc($rankqry);
    $rankqry = $rankqry['rank'];

    $rankupdate = $rankqry + 1;

        if($rankupdate < 0)
        {
        $rankupdate = 0;
        }

    mysql_query ("UPDATE rank SET rank = '$rankupdate' WHERE apt = '$currAdd' ");
    $likesqry = mysql_query ("SELECT * FROM rank WHERE apt = '$currAdd' ");
    $likesqry = mysql_fetch_assoc($likesqry);
    $likesqry = $likesqry['likes'];

    $likesupdate = $likesqry +1;

    mysql_query ("UPDATE rank SET likes = '$likesupdate' WHERE apt = '$currAdd' ");

            echo "<script type='text/javascript' src='jquery.js'></script>";


        echo "<script type='text/javascript'>
                 $(document).ready(function()
    {
    $('#success').animate({width:300},1000);
    $('#success').animate({height:100},1000);
    $('#success').animate({width:300,height:100},1400);
    $('#success').fadeOut(5000);

    }); </script>";

}

}


if ( isset( $_POST['red'] ) )
{
    $currAdd = $_SERVER["REQUEST_URI"];
    $currAdd = substr($currAdd, 0, -9);
    $currAdd = substr($currAdd, 11);

    $ipquery = mysql_query ("SELECT * FROM rate WHERE ip = '$ip' AND apt = '$currAdd' ");
    $ipquery = mysql_fetch_assoc($ipquery);
    $ipquery = $ipquery['ip'];

    if ($ipquery == $ip)
    {
                echo "<script type='text/javascript' src='jquery.js'></script>";


                echo "<script type='text/javascript'>
                 $(document).ready(function()
    {
    $('#fail').animate({width:300},1000);
    $('#fail').animate({height:100},1000);
    $('#fail').animate({width:300,height:100},1400);
    $('#fail').fadeOut(5000);

    }); </script>";


    }
    else
    {

mysql_query ("INSERT INTO rate VALUES ('','$currAdd', '0', '$ip')");    

$rankqry = mysql_query ("SELECT * FROM rank WHERE apt = '$currAdd' ");
$rankqry = mysql_fetch_assoc($rankqry);
$rankqry = $rankqry['rank'];

$rankupdate = $rankqry - 1;

if($rankupdate < 0)
{
    $rankupdate = 0;
}


mysql_query ("UPDATE rank SET rank = '$rankupdate' WHERE apt = '$currAdd' ");

$dislikesqry = mysql_query ("SELECT * FROM rank WHERE apt = '$currAdd' ");
$dislikesqry = mysql_fetch_assoc($dislikesqry);
$dislikesqry = $dislikesqry['dislikes'];

$dislikesupdate = $dislikesqry +1;

    mysql_query ("UPDATE rank SET dislikes = '$dislikesupdate' WHERE apt = '$currAdd' ");

            echo "<script type='text/javascript' src='jquery.js'></script>";



                    echo "<script type='text/javascript'>
                 $(document).ready(function()
    {
    $('#success').animate({width:300},1000);
    $('#success').animate({height:100},1000);
    $('#success').animate({width:300,height:100},1400);
    $('#success').fadeOut(5000);

    }); </script>";


    }
}
?>

我使用了jquery库的1.4.2版本和最新版本，但仍然没有运气。在此先感谢您的帮助。

Answer 1

我正在使用最新版本的jQuery（1.6.1），以下代码在我的服务器上完全正常运行：

<?php

echo '<html>
<head>
</head>
<body>

<div id="success">
    Success! Success! Success! Success! Success! Success! Success! Success! Success! Success! Success! Success! Success! Success!
</div>

<script type="text/javascript" src="js/jquery-1.6.1.min.js"></script>
<script type="text/javascript">
$(document).ready(function()
{
    $(\'#success\').animate({width:300},1000);
    $(\'#success\').animate({height:100},1000);
    $(\'#success\').animate({width:300,height:100},1400);
    $(\'#success\').fadeOut(5000);
});
</script>
</body></html>';

?>

编辑：这是link to my test PHP script

第二次编辑：我想在我的回答中添加一些内容。没有上下文，很难判断这是否是一个问题，但确保在对jQuery进行任何调用之前加载jQuery JS文件。您可以在HTML块之外完成动画调用，但在加载jQuery JS文件之前无法完成。例如，这将起作用：

<?php

echo '<html>
<head>
</head>
<body>

<div id="success">
    Success! Success! Success! Success! Success! Success! Success! Success! Success! Success! Success! Success! Success! Success!
</div>

<script type="text/javascript" src="js/jquery-1.6.1.min.js"></script>
</body></html>';

if ( isset( $_POST['submit'] ) )
{
    mysql_query ("INSERT INTO table VALUES ('','', '', '')");

    echo "<script type='text/javascript'>
             $(document).ready(function()
    {
        $('#success').animate({width:300},1000);
        $('#success').animate({height:100},1000);
        $('#success').animate({width:300,height:100},1400);
        $('#success').fadeOut(5000);
    }); </script>";
}

?>

但这不会：

<?php

echo '<html>
<head>
</head>
<body>

<div id="success">
    Success! Success! Success! Success! Success! Success! Success! Success! Success! Success! Success! Success! Success! Success!
</div>

<script type="text/javascript">
$(document).ready(function()
{
    $(\'#success\').animate({width:300},1000);
    $(\'#success\').animate({height:100},1000);
    $(\'#success\').animate({width:300,height:100},1400);
    $(\'#success\').fadeOut(5000);
});
</script>
<script type="text/javascript" src="js/jquery-1.6.1.min.js"></script>
</body></html>';

?>

使用在线jQuery 1.6.1文件

<form action= '<?php echo $_SERVER['PHP_SELF']; ?>'; method='post'>
<table align='left' width='500px' border='0' cellpadding='0' cellspacing='0'>
<?
   // Using online jQuery 1.6.1 file
   echo"      
      <tr>
        <td>
            <div id='blue'><input type='image' name='blue' id='likebutton' value='blue' src='../zrating/like.png' 

            /></div>

        </td>
        <td>
        <b>$numlikes Like</b>
        <div id='like' style='width:$likewidth'><div id='dis'></div></div><br>
        </td>
      </tr>
      <tr>
        <td>
            <div id='red'><input type='image' name='red' id='dislikebutton' value='red' src='../zrating/dislike.png'  

            /></div>

        </td>
        <td>
            <b>$numdislikes Dislike</b>
            <div id='dislike' style='width:$dislikewidth'><div id='dis'></div></div><br>
        </td>
      </tr>
      <tr>
      <td>
      </td>
      <td>
      <div id='success'><br><br>Thanks for voting!</div>
      <div id='fail'><br><br>You are only allowed 1 vote per apartment.</div>     
      </td>
      </tr>
  </table>
</form>
";

echo "<script type='text/javascript' src='http://code.jquery.com/jquery-1.6.1.min.js'></script>";


                echo "<script type='text/javascript'>
                 $(document).ready(function()
    {
    $('#fail').animate({width:200},1000);
    $('#fail').animate({height:100},1000);
    $('#fail').fadeOut(5000);

    }); </script>";
?>

Answer 2

如果它适用于除Firefox 4以外的所有浏览器，您可能需要查看此主题：

http://forum.jquery.com/topic/jquery-1-4-x-firefox-4-beta-ajax-callbacks

是否使用ajax提交数据？