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-->
答案 0 :(得分:2)
指定classNames
代替id's
然后使用此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
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-->