我正在测试以下html表单。
<html>
<head>
<link href = "style.css" rel = "stylesheet" type = "text/css">
</head>
<form action = "test1.php" enctype = "multipart/form-data" method = "POST">
<input type = "hidden" name = "playno" value = "testing">
<input type = "image" src = "uploads/defb.png" name = "submit" value = "submit"/>
</form>
</html>
以下内容保存在“test1.php”中:
<?php
$hiddenvalue = $_POST['playno'];
if (isset($_POST['submit'])){
echo "OK";
}
else
{
echo "error";
}
?>
在broswer中,当我打印$ hiddenvalue时,我返回值“testing”。但是,每次输出“错误”,而不是“确定”。
我非常感谢任何帮助。这让我发疯! 非常感谢提前。
答案 0 :(得分:2)
使用input type="image"
时,浏览器会发送submit_x
和submit_y
。
因此在PHP中,$_POST['submit']
将无法使用,但会定义$_POST['submit_x']
和$_POST['submit_y']
(包含点击图像的X / Y坐标)。
答案 1 :(得分:1)
只需在表单中添加:
<input type="hidden" name="submit" value="submit"/>
这样它就会在你的PHP中收到$_POST['submit']
。
答案 2 :(得分:-1)
<input type = "image" src = "uploads/defb.png" name = "submit" value = "submit"/>
更改为:
<input type = "submit" name = "submit" value = "submit"/>
因为输入没有这种类型
创建输入样式 HTML
<input type = "submit" name = "submit" value = "submit" id="button" />
CSS
#button {
background:url("uploads/defb.png");
border:0;
outline:0;
}