【发布时间】:2019-02-25 10:55:24
【问题描述】:
(第一页)
<form action="page1.php" method="post">
<input type="image" src="images/leavecredits.png" name="btnImage" id="btnImage" value="uuuu"/>
<input type="hidden" id="hiddenEmpID" name="hiddenEmpID" value="yyyy"/>
<input type="image" src="images/leavecards.png" alt="Submit" width="100" height="100" style="background-color:#FFF; border:dotted #CCCCCC thin;" onMouseOver="this.src='images/leavecards_onmouseover.png'" onMouseOut="this.src='images/leavecards.png'"
id="btnLeaveManagement2" name="btnLeaveManagement2" value="leavecards">
</form>
(page1.php)
<?PHP
echo "<br>hiddenEmpID: " . $_POST['hiddenEmpID'];
echo "<br>btnImage: " . $_POST['btnImage'];
?>
结果:
hiddenEmpID: yyyy
注意:未定义索引:btnImage .... in line....
btnImage:
图像输入类型的值不会出现在下一页 浏览器:chrome 和 mozilla
你们觉得哪里不对?
(如果 value 真的不起作用,可能的解决方案) 大家好,因为我真的想尝试使用 type="image" 作为按钮,所以我在 php 上想出了这个脏代码
<form action="page1.php" method="post">
<input type="image" src="images/leavecredits.png" alt="Submit" id="btnLeaveManagement" name="btnLeaveManagement" width="100" height="100" style="background-color:#FFF; border:dotted #CCCCCC thin;" onMouseOver="this.src='images/leavecredits_onmouseover.png'" onMouseOut="this.src='images/leavecredits.png'" >
<input type="hidden" id="hiddenEmpID" name="hiddenEmpID" value="leavecards"/>
</form>
<form action="page1.php" method="post">
<input type="image" src="images/leavecredits.png" alt="Submit" id="btnLeaveManagement" name="btnLeaveManagement" width="100" height="100" style="background-color:#FFF; border:dotted #CCCCCC thin;" onMouseOver="this.src='images/leavecredits_onmouseover.png'" onMouseOut="this.src='images/leavecredits.png'" >
<input type="hidden" id="hiddenEmpID" name="hiddenEmpID" value="leavecredits"/>
</form>
echo "<br>hiddenEmpID: " . $_POST['hiddenEmpID'];
现在它将取决于隐藏输入的值
【问题讨论】: