当用户点击按钮时,我想要打开一个弹出窗口,当我点击弹出窗口中的“接受”按钮时,弹出窗口将关闭。但是当它关闭时,我想更改用户点击的按钮以打开弹出窗口。我使用的是图像而不是按钮。请告知我该怎么做。
它不适合我。我认为你没有得到我的问题。“我点击图像然后弹出窗口打开后再次接受该弹出窗口中的whaterver它将显示前一个窗口,然后该图像应该改变”我该怎么办?那样做?我在这里发送更新的代码:
<?php
session_start();
if($_SESSION['typeimagedis']=='image/gif')
$namedis=$_SESSION['typeimagenamedis'];
else
$namedis=$_SESSION['typeimagenamedis'];
if ($_POST['hidcropimg'] == 'hidcropimg')
{
if(file_exists('images/'.$_SESSION['typeimagenamedis']))
{
unlink('images/'.$_SESSION['typeimagenamedis']);
}
$targ_w =$_POST['w'];$targ_h = $_POST['h'];
$jpeg_quality = 20;
if(($_POST['typeimage'] == "image/gif"))
{
$src='upload/'.$_SESSION['typeimagenamedis'];
$img_r = imagecreatefromgif($src);
}
if(($_POST['typeimage'] == "image/jpg")|| ($_POST['typeimage'] == "image/jpeg")|| ($_POST['typeimage'] == "image/pjpeg"))
{
$src='upload/'.$_SESSION['typeimagenamedis'];
$img_r = imagecreatefromjpeg($src);
}
$dst_r = ImageCreateTrueColor( $targ_w, $targ_h );
@imagecopyresampled($dst_r,$img_r,0,0,$_POST['x'],$_POST['y'],
$targ_w,$targ_h,$_POST['w'],$_POST['h']);
$yes=1;
if(($_POST['typeimage'] == "image/gif"))
{
$name=$_SESSION['typeimagenamedis'];
imagegif($dst_r,'images/'.$name,$jpeg_quality);
}
if(($_POST['typeimage'] == "image/jpg")|| ($_POST['typeimage'] == "image/jpeg") || ($_POST['typeimage'] == "image/pjpeg"))
{
$name=$_SESSION['typeimagenamedis'];
imagejpeg($dst_r,'images/'.$name,$jpeg_quality);
}
}
if($_POST['subacc']=='Accept')
{
copy('images/'.$_SESSION['typeimagenamedis'],'crop/'.$_SESSION['typeimagenamedis']);
setcookie ("cronimagename", $_SESSION['typeimagenamedis'], time()+60*60*24*30,'/','iphoneid.com');}
?>
<script type='text/javascript' src='https://ajax.googleapis.com/ajax/libs/jquery/1.7.2/jquery.min.js'></script>
<script type='text/javascript'>
function closewin(){
window.close();
$('#openimg',opener.document).attr('src','/v/vspfiles/photoutility/images/clickheretoTHK_off.gif');
//alert('openimg');
}
</script>
<?php
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN"
"http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<script src="../js/jquery.min.js"></script>
<script src="../js/jquery.Jcrop.js"></script>
<link rel="stylesheet" href="../css/jquery.Jcrop.css" type="text/css" />
<link rel="stylesheet" href="demo_files/demos.css" type="text/css" />
<? if($_GET['ref']==1){?>
<META HTTP-EQUIV="refresh" CONTENT="0; URL=http://serverl.iphoneid.com/demos/croped.php?ProductCode=<?=$_GET['ProductCode']?>">
<? }?>
</head>
<body>
<div id="outer">
<div class="jcExample" style="width: 802px;height: 1344px;">
<table width="151" align="center" style="padding-right:65px;" >
<tr><td width="54"><form id="frmacc" name="frmacc" method="post" action=""><input type="submit" id="subacc" name="subacc" value="Accept" onClick="closewin();" /></form></td><td width="238">
<form id="frmcancel" name="frmcancel" method="post" action="crop.php?ref=1&ProductCode=<?=$_GET['ProductCode']?>" ><input type="submit" id="subcancel" name="subcancel" value="Select Again" /></form></td></tr></table>
<div class="article" style="position:relative; margin:70px 0 0 230px;">
<? if($_GET['ref']==1){
?>
<img src="wait.gif" width="200px" height="75px" >
<?
}else{ ?>
<div style="float:left; height:530px; padding:0px 0 0 0px; width:580px; background:url(<? if($_GET['ProductCode']=='IPBB'){?>black.png<? }elseif($_GET['ProductCode']=='IPBR'){?>red.png<? }elseif($_GET['ProductCode']=='IPBW'){?>white.png<? }?>) no-repeat; position:relative; z-index:999999999999">
</div><!--style="padding-top:<?=$padding?>px"-->
<div align="center" style=" height: 148px; width: 148px; position:absolute; top:161px; left:61px; z-index:1" ><img src="images/<?=$namedis?>" height="148" width="148" /></div>
<? }?>
</div>
</div>
</div>
</body>
</html>
答案 0 :(得分:1)
要从弹出窗口更改按钮的背景图像,请使用以下命令:
<script>
$("#openimg1",opener.document).attr("src","/image-path-here");
</script>