php - 如何将php值传递给jquery并获取

Question

我在传递值时遇到一个问题，无法通过查询字符串传递它。我的 PHP 是:

 <div class="menu">
        <div class="sub_menu_header">ABOUT</div>
        <?php 
            $query1="SELECT id,title FROM aboutus_tbl";
            $resulto=mysql_query($query1);
            while ($row = mysql_fetch_array($resulto)) {
        ?>
        <a href="about" class="menuid" id="<?php echo $row['id'];?>"><div class="<?php echo $row['title'];?>" id="sub_menu"><?php echo $row['title'];?>
        </div></a>
        <?php
        }
        ?>
    </div>

jQuery:

 <script>
    $(".menuid").click(function() {
        date_time = $(this).attr('id');
        console.log(data);
        $.ajax({
            type: 'POST',
            url: 'about',
            data: {"aboutus_id" : date_time},
            success: function(data){
                alert( data );
            }
        });
    });
    </script>

完整代码:

<?php ob_start(); include('web/header.php');?>
<!--main-->
<div class="main_btm">
<div class="wrap">
<div class="main">
    <div class="new_head"><font size="5" color="#FC2B5F">
    <script>
    $(".menuid").click(function() {
        date_time = $(this).attr('id');
        console.log(data);
        $.ajax({
            type: 'POST',
            url: 'about',
            data: {"aboutus_id" : date_time},
            success: function(data){
                alert( data );
            }
        });
    });
    </script>
    <?php
    if (isset($_POST['aboutus_id'])) {
        $id=$_POST['aboutus_id'];
        echo $id;
    }
    else{
        $id=1;
    }
    $que="SELECT title FROM aboutus_tbl WHERE id='".$id."'";
    $re=mysql_query($que);
    $r=mysql_fetch_array($re);
    echo $r[0];
    ?>
    </font></div>
    <div class="menu">
        <div class="sub_menu_header">ABOUT</div>
        <?php 
            $query1="SELECT id,title FROM aboutus_tbl";
            $resulto=mysql_query($query1);
            while ($row = mysql_fetch_array($resulto)) {
        ?>
        <a href="about" class="menuid" id="<?php echo $row['id'];?>"><div class="<?php echo $row['title'];?>" id="sub_menu"><?php echo $row['title'];?>
        </div></a>
        <?php
        }
        ?>
    </div>
    <div class="contant">
    <div class="con_details">
    <?php
        $query="SELECT * FROM aboutus_tbl WHERE id='".$id."'";
        $res=mysql_query($query);
        $result=mysql_fetch_array($res);
        $d1_per=explode("  ", $result['desc1']);
        $d1_lenth=sizeof($d1_per);
        $d2_per=explode("  ", $result['desc2']);
        $d2_lenth=sizeof($d2_per);
        $d3_per=explode("  ", $result['desc3']);
        $d3_lenth=sizeof($d3_per);
        if ($result['sub_title1']) {
    ?>
    <legend><font size="5"><?php echo $result['sub_title1'];?></font></legend>
    <?php
    }
    if ($result['desc1']) {
    for ($i=0; $i < $d1_lenth; $i++) { 
        echo "<p>".$d1_per[$i]."</p>";
    }
    }
    if ($result['sub_title2']) {
    ?>
    <legend><font size="5"><?php echo $result['sub_title2'];?></font></legend>
    <?php
    }
    if ($result['desc2']) {
    for ($i=0; $i < $d2_lenth; $i++) { 
        echo "<p>".$d2_per[$i]."</p>";
    }
    }
    if ($result['sub_title3']) {
    ?>
    <legend><font size="5"><?php echo $result['sub_title3'];?></font></legend>
    <?php
    }
    if ($result['desc3']) {
    for ($i=0; $i < $d3_lenth; $i++) { 
        echo "<p>".$d3_per[$i]."</p>";
    }
    }
    ?>
    </div>
    </div>
</div>
</div>
</div>
<script type="text/javascript">
window.onload = function() {
    $(document).ready(function() {
        $('.<?php echo $result['title']?>').addClass('sub_menu_selected');
    });
}
</script>
<?php include('web/footer.php'); ob_flush();?>

在上面的代码中，我在<div>中显示了一个菜单。和<a>标签在外面，这样我就可以点击它。我的问题是，当我单击菜单时，它必须显示 id menuid的类但它不工作并且 $_POST['aboutus_id']没有获得任何值(value)。

Answer 1

试试这个代码:-如果您想重定向用户:

header('Location: http://ashvin.com/page.php?' . http_build_query($_GET, '', '&')); die();

如果您只想获取页面，请使用:

file_get_contents('http://ashvin.com/page.php?' . http_build_query($_GET, '', '&'));

还有另一个:-

<a href="/exsomeurl.php?id=1&name=Jose" class="ajax-link"> Click </a>
<a href="/exsomeurl.php?id=2&name=Juan" class="ajax-link"> Click </a>
<a href="/exsomeurl.php?id=3&name=Pedro" class="ajax-link"> Click </a>
...
<a href="/exsomeurl.php?id=n&name=xxx" class="ajax-link"> Click </a>

<script type="text/javascript">
$(function() {
   $('.ajax-link').click( function() {
         $.get( $(this).attr('href'), function(msg) {
              alert( "Data Saved: " + msg );
         });
         return false; // don't follow the link!
   });
});
</script>