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我现在有点卡住了,可以使用一些帮助。PHP - 多种形式,在提交时从两者获取值?
我有正在使用由多个部门在进入他们的日程安排每个用户都被要求完成1X选择和2x输入字段被分为FORM 1和FORM 2
FORM 1:
- 部门代码(提交的onclick事件)
FORM 2:
- 名
- 日期
- 提交按钮
用户选择他/她的部门代码后,日历将刷新并获取条目这一特定部门只(这意味着它会过滤所有记录部门代码)。之后,用户必须输入他的用户名并选择日期,然后按提交按钮(表2)。
现在,问题是,为了FORM 2正确提交,我需要知道表单1的部门代码。此外,我有麻烦,防止日历刷新(表单1)时按下提交按钮对于表单2.
总之,在访问这两个表单的数据时,我怎样才能清楚地区分表单1和表单2之间的($ _POST)?
PHP
<?php
// SHOULD BE EXECUTED AFTER THE SUBMIT BUTTON WAS CLICKED
if ($_SERVER['REQUEST_METHOD'] == 'POST' && isset($_POST['type']) && ($_POST['type']=='new_entry')) {
// include connection details
include '../../plugins/MySQL/connect_db.php';
// Open a new connection to the MySQL server
$con = new mysqli($dbhost,$dbuser,$dbpass,$dbname);
// Output any connection error
if ($con->connect_error) {
die('Error : ('. $con->connect_errno .') '. $con->connect_error);
}
// define variables
$table = 'calendar';
$type = mysqli_real_escape_string($con,$_POST['type']);
$holidex = mysqli_real_escape_string($con,$_POST['holidex']);
if($type == 'new_entry')
{
// define variables and query
$mod_property = mysqli_real_escape_string($con,$_POST['holidex']);
$mod_name = mysqli_real_escape_string($con,$_POST['mod_name']);
$mod_date = date('Y-m-d',strtotime($_POST['mod_date']));
$sql = "INSERT INTO calendar (`title`, `startdate`, `enddate`, `allDay`, `color`, `holidex`) VALUES ('$mod_name','$mod_date','$mod_date','true','','$mod_property')";
print($sql);
$result = $con->query($sql) or die('<p>Could not submit new MOD record into database: ' . MYSQLI_ERROR() . '</p>');
$result->free();
}
$con->close();
}
?>
<form name="form1" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
<!-- <form name="mod_form" action="../../plugins/MySQL/ajax_action.php?type=new_entry" method="POST"> -->
<!-- Property -->
<div class="col-md-4">
<label>Property</label>
<div class="input-group">
<span class="input-group-addon"><i class="fa fa-server"></i></span>
<select name="holidex" id="holidex" class="form-control select2" style="width: 100%;" data-placeholder="Select your property" onchange="this.form.submit();" method="POST" <?php if($_SESSION['Access']=='User') { echo "disabled"; } ?>>
// get all my departments and their respective codes
<option value="1">Dept. 1</option>
<option value="2">Dept. 2</option>
<option value="3">Dept. 3</option>
</select>
</div>
<!-- /btn-group -->
</div>
<!-- /.property -->
</form>
<form name="form2" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
<!-- MOD Name -->
<div class="col-md-4">
<label>MOD Name</label>
<div class="input-group">
<span class="input-group-addon"><i class="fa fa-user"></i></span>
<select name="mod_name" id="mod_name" class="form-control select2" style="width: 100%;" data-placeholder="Select a name">
<option value=""></option>
<option value="1">User1</option>
<option value="2">User2</option>
<option value="3">User3</option>
</select>
</div>
<!-- /.input-group -->
</div>
<!-- /.mod name -->
<!-- MOD Date -->
<div class="col-md-3">
<label>MOD Date</label>
<div class="input-group">
<span class="input-group-addon"><i class="fa fa-server"></i></span>
<input type="date" class="form-control" name="daterangepicker" id="daterangepicker" />
<!-- <input type="date" id="mod_date" name="mod_date" class="form-control" style="width: 100%;"> -->
</div>
</div>
<!-- /.mod date -->
// hidden input field to determine the type and help differentiate the $_POST submissions
<input type="hidden" class="form-control" name="type" id="type" value="new_entry"/>
<!-- Submit button -->
<div class="col-md-1 text-center">
<label> </label>
<button type="submit" name="btnSubmit" class="btn btn-primary btn-block" onclick="this.disabled=true; this.value = 'Wait...'; this.form.submit(); return true;">Submit</button>
</div>
<!-- /.submit button -->
</form>
<!-- /.form 2 -->
我会用JavaScript来发送表单,让您可以添加额外的数据。或在部门代码 – Sitethief
中用javascript创建表单2中的隐藏字段我的JS很糟糕,但您的意思是沿着表单2创建一个隐藏的JS字段,并且在form1中的部分代码字段发生更改时,更新隐藏字段与输入值,是否正确? – Armitage2k
是,这是更简单的解决方案之一。 – Sitethief