[英]insert multiple rows data into database fetched from another table
我正在尝试在php中建立一个在线出席者,我正在从学生表中获取学生列表,并且想要将所有chkboxs都选中(如果选中),如果未选中,则无法这样做。
<div class="attendance">
<form accept="att.php" method="POST">
<?php
$sel_sql = "SELECT * FROM student";
$run_sql = mysqli_query($conn,$sel_sql);
while($rows = mysqli_fetch_assoc($run_sql)){
$id = $rows['id'];
echo '<input type="checkbox" name="status" value="'.$id.'" checked>'.ucfirst($rows['f_name']).'';
}
?>
<input type="submit" name="submit_attendance" value="Post Attendance">
<?php echo $msg; ?>
</form>
</div>
它显示了完美的学生列表,但我不知道如何为所有这些chkbox设置插入查询
尝试此查询从另一个表插入
SELECT * INTO TABLE2 FROM student
使用学生表上的where
条件作为where student.column值来选择选中的值
将上述内容与复选框输入字段一起应用
echo '<input type="checkbox" name="status" value="'.$id.'" 'if($row['present_field_of_database']) ? 'checked' : ''
'>'.ucfirst($rows['f_name']).'';
没问题,然后用您的问题更新代码,希望对您有用
<div class="attendance">
<form accept="att.php" method="POST">
<?php
$sel_sql = "SELECT * FROM student";
$run_sql = mysqli_query($conn,$sel_sql);
while($rows = mysqli_fetch_assoc($run_sql)){
$id = $rows['id'];
// if your $id value is right from $rows['id'] then
// change your student table name to the another table where available status of the student
$wh_sql = "SELECT * FROM student WHERE id=".$id;
$run_wh_sql = mysqli_query($conn, $wh_sql);
$Wh_rows = mysqli_fetch_assoc($run_wh_sql);
// add student present or absent value to the rows data
$rows['status'] = $Wh_rows['status'];
}
// set value as A or P respectively absent or present add jquery plugins for onclick event while client click on checkbox change value respectively
echo '<input type="checkbox" name="status" value="'.$rows['status'].'" 'if($rows['status'] == "A") ?'checked': '' '>'.ucfirst($rows['f_name']).' onclick= "$(this).val(this.checked ? P : A)"';
?>
<input type="submit" name="submit_attendance" value="Post Attendance">
<?php echo $msg; ?>
</form>
</div>
if (isset($_POST['send'])) {
$s_id = $_POST['status'];
$id = $_POST['student'];
if ($s_id) {
foreach ($s_id as $s) {
foreach ($id as $i) {
if (mysqli_query($conn, "INSERT INTO attendence(s_id,status) VALUES('".
mysqli_real_escape_string($conn, $s)."','".
mysqli_real_escape_string($conn, $i)."')")) {
$msg = "success";
}else{
$msg = "failed";
}
}
}
}
}
我有3个学生。 当我按发送它发送9个条目。 我无法理解如何插入所有学生数据
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