为创建的div分配唯一ID,然后使用jquery引用

时间:2013-06-06 21:09:52

标签: php jquery dynamic-data unique-id

我是php的新手,是jQuery的初学者。我有一个php页面,其中填充了mySQL表中的数据。我想要实现的是包含“查看作业”的h3具有分配给它的唯一ID,以及打印出作业描述的div。然后,我想用jQuery引用它们,这样如果有人单击View Job,则只显示该作业的描述。我希望这是有道理的。

我尝试使用类,当然,当点击任何View Job时,所有作业描述都会显示出来。

我尝试了solution here,但最终我的页面上有36个“查看作业”链接,我需要在创建时为h3和div分配唯一ID。

我愿意接受另一种方法来实现我正在寻找的东西 - 基本上是在用户点击每个作业的“查看作业”时隐藏/折叠每个描述。

这是我的代码:

        <script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
        <style type="text/css">
        .job-post{border-bottom: 1px solid red; padding: 0; margin: 10px 0;}
        h3, p{padding: 0; margin:0;}
        .view-job{cursor: pointer;}
        </style>
        <script type="text/javascript">
                        $(document).ready(function() {
                            $("div#job-details").css("display" , "none");
                             $("h3#view-job").click(function() {
                                $("div#job-details").css("display" , "block");
                            });

                        });
         </script> 
        <?php 
         // Connects to your Database 
         mysql_connect("xx", "xx", "xx") or die(mysql_error()); 
         mysql_select_db("lcwebsignups") or die(mysql_error()); 
         $data = mysql_query("SELECT * FROM openjobs") 
         or die(mysql_error()); 
          ?>
         <div id="job-container"> 
          <?php 
         Print ""; 
         while($info = mysql_fetch_array( $data )) 
         { 
         Print "<div class='job-post'>"; 
         Print "<h3>Position / Location:</h3> <p>".$info['jobtitle'] . ", ".$info['joblocation'] . "</p>";
         Print "<h3 id='view-job'>View Job:</h3> <div id='job-details'>".$info['jobdesc'] . " </div>"; 
         Print "</div>";
         }  
         ?>
        </div><!--//END job-container-->

3 个答案:

答案 0 :(得分:2)

指定classNames代替id's

然后使用上下文选择仅在上下文中搜索的div而不是所有div

$("h3.view-job").on('click',function() {
    $(this).next("div.job-details").css("display" , "block");
});

答案 1 :(得分:0)

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
<style type="text/css">
.job-details{display:none;} 
.job-post{border-bottom: 1px solid red; padding: 0; margin: 10px 0;}
h3, p{padding: 0; margin:0;}
.view-job{cursor: pointer;}
</style>
<script type="text/javascript">
$(document).ready(function() {
     // Please hide this class on CSS
     // $("div.job-details").css("display" , "none");
     $("h3.view-job").click(function() {
        // you ned to use this keyword
        // it is very important when you use javascript or jquery
        // this keyword only pick element on which you click
        $(this).parent().find(".job-details").show();
    });

});
   </script> 
    <?php 
     // Connects to your Database 
     mysql_connect("xx", "xx", "xx") or die(mysql_error()); 
     mysql_select_db("lcwebsignups") or die(mysql_error()); 
     $data = mysql_query("SELECT * FROM openjobs") 
     or die(mysql_error()); 
     ?>
     <div id="job-container"> 
     <?php 
     // First of all Please use echo instead of Print
     echo ""; 
     while($info = mysql_fetch_array( $data )) 
     { 
         echo "<div class='job-post'>"; 
         echo "<h3>Position / Location:</h3> <p>".$info['jobtitle'] . ", ".$info['joblocation'] . "</p>";
         // you can't use id multiple time in a same page so instead of id use Class
         // if you want to use id then you have to generate uniq id
         // check below I generate id --> I don't know your database field that's why I used jobId
         echo "<h3 class='view-job' id='job".$info['jobId']."'>View Job:</h3> <div class='job-details'>".$info['jobdesc'] . " </div>"; 
         echo "</div>";
     }  
     ?>
    </div><!--//END job-container-->

答案 2 :(得分:0)

这应该做你想要的(测试好了)。只需替换你的mysql登录信息,这个例子就可以了。

我完成了你的工作细节div(它缺少来自php的数据),但你真正想要的是jQuery代码。

如果点击了以<h3> 开头的view-job标记,则jQuery点击事件处理程序将:

  • 重新隐藏所有div ,其ID以字符job-details

  • 开头
  • 显示DOM树中的下一个div


CODE:

<script src="http://ajax.googleapis.com/ajax/libs/jquery/1.8.3/jquery.min.js"></script>
<style type="text/css">
    .job-post{border-bottom: 1px solid red; padding: 0; margin: 10px 0;}
    h3, p{padding: 0; margin:0;}
    .view-job{cursor: pointer;}
</style>
<script type="text/javascript">
    $(document).ready(function() {
        $("div[id^='job-details']").css("display" , "none");

        $('h3[id^="view-job"]').click(function() {
            $("div[id^='job-details']").hide(500);
            $(this).next('div').show(500);
        });

    });
 </script> 

 <?php 
    // Connects to your Database 
    mysql_connect("xx", "xx", "xx") or die(mysql_error()); 
    mysql_select_db("lcwebsignups") or die(mysql_error()); 
    $data = mysql_query("SELECT * FROM openjobs") or die(mysql_error()); 

        $data = mysql_query("SELECT * FROM openjobs") or die(mysql_error());
  ?>

<div id="job-container"> 

<?php 
    echo "<br>"; 
    while($info = mysql_fetch_assoc( $data )) { 
        echo "<div class='job-post'>"; 
        echo "<h3>Position / Location:</h3> <p>".$info['jobtitle'] . ", ".$info['joblocation'] . "</p>";
        echo "<h3>View Job:</h3> <div id='job-details'>".$info['jobdetails']."</div>"; 
        echo "</div>";
    }
 ?>

 </div><!--//END job-container-->