Question

我很难对复选框进行验证，但似乎无法理解它出错的地方。任何帮助都会很棒。因为它作为一个数组从MySQL中拉出来，我不确定这是否是我错的地方。

<script language="JavaScript" type="text/javascript">
function checkform ( form )
{
  if(!form.check_box_delete[].checked) {
        alert('Please check at least one of the options.');
        return false;
    }
    return true;
}
</script>

以下是正文的代码：

<form name="form" method="post" action="loan_checkedIn.php" onsubmit="return checkform(this);">
<?php
echo "<table border='1' align='center' width='900'>
<tr>
<th bgcolor='#00a3e0'><font face='Arial'>Select</font></th>
<th bgcolor='#00a3e0'><font face='Arial'>Customer</font></th>
<th bgcolor='#00a3e0'><font face='Arial'>Make</font></th>
<th bgcolor='#00a3e0'><font face='Arial'>Model</font></th>
</tr>";

$results = mysql_query("SELECT * FROM loan WHERE email='$email' AND status='Out'");
while($row1 = mysql_fetch_array($results))
  {
  echo "<tr>";
  echo "<td align='center'><input type='checkbox' name='check_box_delete[]' value='" . $row1['id'] . "'></td>";
  echo "<td align='center'><font face='Arial'>" . $row1['customer'] . "</font></td>";
  echo "<td align='center'><font face='Arial'>" . $row1['make'] . "</font></td>";
  echo "<td align='center'><font face='Arial'>" . $row1['model'] . "</font></td>";
  echo "</tr>";
  }
  echo "</table>";
  ?>

  <p align="center"><input type="submit" style="background-color:#00A3E0; color:#FFFFFF;" name="submit" value="Return Equipment"></p>
  </form>

Answer 1

您可以遍历表单中的所有元素i。

Answer 2

你在javascript中有问题。更改您的代码如下：

<script language="JavaScript" type="text/javascript">
function checkform ( form )
{  
  var checkboxs=form['check_box_delete[]'];
  var cheked=false;
  for(var i=0,l=checkboxs.length;i<l;i++)
  {
      if(checkboxs[i].checked)
      {
          cheked=true;
          break;
      }
  }
  if(!cheked)
  {
    alert('Please check at least one of the options.');
        return false;
  }  
    return true;
}
</script>

也停止使用deprected / removed mysql _ 扩展名。使用 mysqli_ 或 PDO （使用准备好的查询）

Javascript复选框验证不能在PHP中工作

2 个答案: