【发布时间】:2018-03-27 17:22:53
【问题描述】:
我正在尝试使用基于 buttonclick 的表中的数据填充表单。我正在将数据传递给 javascript 函数。
<?php
$sql=mysqli_query($con, "SELECT FacultyEducationalId, FacultyDegreeName, FacultyCollegeName, FacultyPassingYear FROM facultyeducationaltable WHERE FacultyId=".$_SESSION['userid']);
while ($row=mysqli_fetch_array($sql))
{
?>
<tr class="info">
<td><?php echo $row['FacultyDegreeName'] ?></td>
<td><?php echo $row['FacultyCollegeName'] ?> </td>
<td><?php echo $row['FacultyPassingYear'] ?> </td>
<td>
<button onclick="modifyEdu(<?php echo $row['FacultyEducationalId'].','.$row['FacultyDegreeName'].','.$row['FacultyCollegeName'].','.$row['FacultyPassingYear']; ?>)" id="modifyEdu" class="btn btn-primary">Modify</button>
</td>
</tr>
<?php
}
?>
函数如下:
<script type="text/javascript">
function modifyEdu(id, deg, uniname, passyear) {
var facultyeduid=parseInt(id);
var degname=deg;
var uniname=uniname;
var passyear=passyear;
console.log("1");
document.getElementById("degreename").value="2";
document.getElementById("universityname").value=uniname;
document.getElementById("passyear").value=passyear;
}
</script>
我得到:未捕获的 ReferenceError:x 未在 HTMLButtonElement.onclick 中定义。我该如何解决? (x 是 'FacultyDegreeName' 中的值)
【问题讨论】:
-
好吧,你不要用引号括起来听起来像。查看源代码,您就会明白为什么。
标签: javascript php mysqli