<html>
<input type="hidden" name="username[]" value="<?php echo $userRow['username']; ?>" readonly />
<input type="hidden" name="school[]" value="<?php echo $userRow['school']; ?>" readonly />
<SELECT name='cand_name[]' class='form-control'>
<option></option>
<option>Ahmed MUSA</option>
<option>Ahmed Arinze</option>
<option>NOsa Igiebor</option>
</SELECT>
<SELECT name='cand_name[]' class='form-control'>
<option></option>
<option>Ahmed MUSA</option>
<option>Ahmed Arinze</option>
<option>NOsa Igiebor</option>
</SELECT>
</html>
以下代码;
<?php
include_once 'dbcon.php';
{
$username = $_POST['username'];
$cand_name = $_POST['cand_name'];
$school = $_POST['school'];
for ($i = 0; $i < count($username); $i++) {
$username = ($username[$i]);
$cand_name = ($cand_name[$i]);
$school = ($school[$i]);
mysqli_query($con, "INSERT INTO parlia (username, cand_name, school) VALUES ('$username', '$cand_name', '$school')");
}
}
我检查了其他SO答案,但似乎没有一个对我有用。 我已经编辑了它仍然无法正常工作。当我使用输入尝试此代码时,它工作得很好,但不想使用选项
答案 0 :(得分:0)
<?php echo $userRow['username']; ?>
和<?php echo $userRow['school']; ?>
来自哪里?这来自dbase表中的字段..只有登录用户才能访问我正在处理的页面
答案 1 :(得分:0)
这段代码对我有用..但是你很容易受到sql注入
<?php
$username = $_POST['username'][0];
$school = $_POST['school'][0];
$candname = $_POST['candname'];
foreach ($_POST['candname'] as $candname) {
$query = sprintf(
"INSERT INTO parlia_votes (username, school, candname) VALUES ('%s', '%s', '%s')",
$username,
$school,
$candname
);
$con->query($query);
}
?>