javascript onclick更改图像无法正常工作
我有两个数据,每个数据有4个图像,我想一个接一个地显示,我想通过使用onclick javascript单击按钮来交换图像。但它并不适用于所有数据,只适用于数据的末尾。这是我的密码javascript onclick更改图像无法正常工作,javascript,onclick,Javascript,Onclick,我有两个数据,每个数据有4个图像,我想一个接一个地显示,我想通过使用onclick javascript单击按钮来交换图像。但它并不适用于所有数据,只适用于数据的末尾。这是我的密码 <?php include "koneksi.php"; // connect to database ?> <html> <head> <title></title> <style> img{ height:100px
<?php
include "koneksi.php"; // connect to database
?>
<html>
<head>
<title></title>
<style>
img{
height:100px;
width:100px;
}
</style>
</head>
<body>
<table>
<?php
$result=mysqli_query($koneksi, "SELECT * FROM tbbarang where idbarang='43' or idbarang='44'");
$i= 1;
while ($row=mysqli_fetch_array($result))
{ ?>
<tr>
<!--image 1--> <td><img id="img<?php echo $i?>" src=" <?php echo $row['gmbr1'] ?>"></td> <!-- image that i want to swap with folowing images 1, image 2, image 3, image 4-->
<!--image 2--> <td><img src="<?php echo $row['gmbr2'] ?>"></td>
<!--image 3--> <td><img src="<?php echo $row['gmbr3'] ?>"></td>
<!--image 4--> <td><img src="<?php echo $row['gmbr4'] ?>"></td>
<script>
function changeImage(id) {
var image = document.getElementById("img"+id);
if (image.src.match("<?php echo $row['gmbr1']?>")) {
image.src = "<?php echo $row['gmbr2']?>";
} else if (image.src.match("<?php echo $row['gmbr2']?>")) {
image.src = "<?php echo $row['gmbr3']?>";
} else if (image.src.match("<?php echo $row['gmbr3']?>")) {
image.src = "<?php echo $row['gmbr4']?>";
} else if (image.src.match("<?php echo $row['gmbr4']?>")) {
image.src = "<?php echo $row['gmbr1']?>";
}
}
</script>
<td><button id="ubah" onClick="changeImage(<?php echo $i?>)">Click To Change</button></td>
</tr>
<?php
$i=$i+1;
}
?>
</table>
</body>
</html>
您的文件可能不包含任何像这样更改代码的changeimage函数
<?php
include "koneksi.php"; // connect to database
?>
<html>
<head>
<title></title>
<style>
img{
height:100px;
width:100px;
}
</style>
<script>
function changeImage(id, gmbr1, gmbr2, gmbr3, gmbr4) {
var image = document.getElementById("img"+id);
if (image.src == gmbr1) {
image.src = gmbr2;
} else if (image.src == gmbr2) {
image.src = gmbr3;
} else if (image.src == gmbr3) {
image.src = gmbr4;
} else if (image.src == gmbr4) {
image.src = gmbr1;
}
}
</script>
</head>
<body>
<table>
<?php
$result=mysqli_query($koneksi, "SELECT * FROM tbbarang where idbarang='43' or idbarang='44'");
$i= 1;
while ($row=mysqli_fetch_array($result))
{
$gmbr1 = $row['gmbr1'];
$gmbr2 = $row['gmbr2'];
$gmbr3 = $row['gmbr3'];
$gmbr4 = $row['gmbr4'];
?>
<tr>
<!--image 1--> <td><img id="img<?php echo $i?>" src=" <?php echo $gmbr1 ?>"></td> <!-- image that i want to swap with folowing images 1, image 2, image 3, image 4-->
<!--image 2--> <td><img src="<?php echo $gmbr2 ?>"></td>
<!--image 3--> <td><img src="<?php echo $gmbr3 ?>"></td>
<!--image 4--> <td><img src="<?php echo $gmbr4 ?>"></td>
<td><button id="ubah" onClick="changeImage("<?php echo $i?>", "<?php echo $gmbr1 ?>", "<?php echo $gmbr2 ?>", "<?php echo $gmbr3 ?>", "<?php echo $gmbr4 ?>")">Click To Change</button></td>
</tr>
<?php
$i=$i+1;
}
?>
</table>
</body>
</html>
您的按钮调用“changeImage”函数的值是否正确?我可以在代码中看到您可能没有changeImage函数。我认为参数应该是字符串类型。立即试用该功能在单击按钮时是否收到呼叫?我可以稍微试用一下您的系统,但仍然没有运行bro。我认为问题在于从其参数发送值。我试图将参数值显示给innerHTML,但它是空的。然后,我可能无法从表中提取数据