我有一个 tbl_employee 表和 tbl_time 表,我想为出勤插入多个数据插入,但是当我单击提交按钮时,它只插入单个数据。但是问题出在哪里,帮我找出这个..这是插入代码
require './db_connect.php';
class Time extends Db_connect {
protected $link;
public function __construct() {
$this->link = $this->database_connection();
}
public function attendance_insert($data) {
extract($data);
$cur_date = date('Y-m-d');
foreach ($time_attendance as $attn_key => $attn_value) {
if ($attn_value == 'P') {
$SQL = "INSERT INTO tbl_time(employee_id,time_date,time_attendance)VALUES('$attn_key','$cur_date','P')";
$atten_date=mysqli_query($this->link, $SQL);
} else if ($attn_value == 'A') {
$SQL = "INSERT INTO tbl_time(employee_id,time_date,time_attendance)VALUES('$attn_key','$cur_date','A')";
$atten_date=mysqli_query($this->link, $SQL);
}
if ($atten_date) {
$massage = "<div class='alert alert-success text-center'><h5>Attendance insert successfully</h5></div>";
return $massage;
} else {
die('Attendance insert query problem' . mysqli_error($this->link));
}
}
}
}
这是html代码
<?php
require_once './time.php';
$obj_time = new Time()
$massage = '';
if (isset($_POST['btn'])) {
$massage = $obj_time->attendance_insert($_POST);
}
$employee_view = $obj_employee->employee_all_view();
?>
<div class="container-fluid">
<div class="row">
<center>
<a href="time.php" class="btn btn-sm btn-default glyphicon glyphicon-backward pull-left" id="show_form"></a>
<span style="font-size:1.8em;">Attendance form</span>
</center>
</div>
</div>
<hr/>
<?php echo $massage; ?>
<div class="container-fluid">
<div class="row">
<div class="col-lg-12">
<div class="panel panel-body panel-default">
<div class="well text-center" style="font-size:15px;">
<strong>Date :</strong>
<?php $current_date = date('Y-M-d');
echo $current_date; ?>
</div>
<form class="form-horizontal" method="post">
<table class="table table-striped table-responsive text-center">
<tr>
<td><b>Serial</b></td>
<td><b>Name</b></td>
<td><b>ID</b></td>
<td><b>Attendance</b></td>
</tr>
<?php
$i = 0;
while ($employee_info = mysqli_fetch_assoc($employee_view)) {
extract($employee_info);
$i++;
?>
<tr>
<td><?php echo $i; ?></td>
<td><?php echo $employee_first_name . ' ' . $employee_last_name; ?></td>
<td><?php echo $employee_id; ?></td>
<td>
<input type="radio" name="time_attendance[<?php echo $employee_id; ?>]" value="P">P
<input type="radio" name="time_attendance[<?php echo $employee_id; ?>]" value="A">A
</td>
</tr>
<?php } ?>
<tr>
<td colspan="4">
<input type="submit" class="btn btn-primary btn-block" name="btn" value="submit"/>
</td>
</tr>
</table>
</form>
</div>
</div>
</div>
</div>
最佳答案
这个:
foreach($time_attendance as $attn_key => $attn_value){
$SQL = "INSERT INTO tbl_time(employee_id,time_date,time_attendance)
VALUES('$attn_key','$cur_date','A')";
$atten_date = mysqli_query($this->link, $SQL);
if ($atten_date) {
$massage = "<div class=''><h5>Attendance insert successfully</h5></div>";
return $massage;
}
}
如果查询成功,直接要求使用 return $massage;
中断 foreach...所以是的,您将只有一个记录。
将您的返回放在 foreach 之外。
关于php - 为什么不在此代码中插入多个数据,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/40961110/