[英]How to insert data in first form and update the same data in second form and so on
我想問一下，如果要在第一次單擊按鈕（位於第一頁）上插入數據，然后在其余表單上使用新字段更新數據庫時該怎么辦？
這是index.php：
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<body>
<div class="wrapperDiv">
<form action="form-page2.php" method="post" id="form">
<table>
<tr>
<th colspan="3" scope="row">
<h3 style="border-bottom:1px solid #000;">Form 1 </h3>
<div align="center"></div></th>
</tr>
<tr>
<th scope="row"><div align="right">Name</div></th>
<td><div align="center"><strong>:</strong></div></td>
<td><input type="text" name="name" class="required" value="" /></td>
</tr>
<tr>
<th scope="row"><div align="right">Email</div></th>
<td><div align="center"><strong>:</strong></div></td>
<td><input type="text" name="email" class="required" value="" /></td>
</tr>
<tr>
<th scope="row"><div align="right">Mobile</div></th>
<td><div align="center"><strong>:</strong></div></td>
<td><input type="text" name="mobile" value="" /></td>
</tr>
<tr>
<th scope="row"> </th>
<td> </td>
<td><input type="submit" name="submitBtn1" id="submitBtn1" value="Next" /></td>
</tr>
</table>
</form>
</div><!-- end of .wrapperDiv -->
</body>
</html>
我在第二個表單中插入了第一頁的數據，但是現在當用戶填寫第二個表單時，應該在同一字段中更新數據。 我要在其中執行更新查詢的第二頁：
<?php
if (isset($_POST['submitBtn1'])) {
$servername = "localhost";
$username = "root";
$password = "";
$database = "phpmultipage";
$conn = mysqli_connect($servername, $username, $password, $database);
if (!$conn) {
die("connection failed: " . mysqli_connect_error());
} else {
echo 'you are on';
}
$name = $_POST['name'];
$email = $_POST['email'];
$mobile = $_POST['mobile'];
$sql = "INSERT into detail (name,email,contact) VALUES ('$name','$email','$mobile')";
if (mysqli_query($conn, $sql)) {
echo 'record added. Your ID:' . mysqli_insert_id($conn);
$id = mysqli_insert_id($conn);
echo $id;
} else {
echo 'error';
}
mysqli_close($conn);
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<title>Form 2</title>
</head>
<body>
<div class="wrapperDiv">
<form action="form-page3.php" method="post" id="form">
<?php
if(!isset($_POST['submitBtn1'])) header('location: index.php');
foreach($_POST as $key => $value) {
if($key != 'submitBtn1') {
?>
<input type="hidden" name="<?php echo $key; ?>" value="<?php echo $value; ?>" />
<?php
}
}
?>
<table width="500">
<tr>
<th colspan="3" scope="row"><h3>Form 2</h3></th>
</tr>
<tr>
<th scope="row"><div align="right">Password</div></th>
<td><div align="center"><strong>:</strong></div></td>
<td><input type="password" id="password" name="password" value="" /></td>
</tr>
<tr>
<th scope="row"><div align="right">Confirm Password</div></th>
<td><div align="center"><strong>:</strong></div></td>
<td><input type="password" name="confirm_password" id="confirm_password" value="" /></td>
</tr>
<tr>
<th scope="row"> </th>
<td> </td>
<td><input type="submit" name="submitBtn3" id="submitBtn3" value="Next" /></td>
</tr>
</table>
</form>
</div>
</body>
</html>
為了執行此操作，我在第3頁上寫道：
<?php
if (isset($_POST['submitBtn3'])) {
$servername = "localhost";
$username = "root";
$password = "";
$database = "phpmultipage";
$con = mysqli_connect($servername, $username, $password, $database);
if (!$con) {
die("connection failed: " . mysqli_connect_error());
} else {
echo 'you are on';
}
$password = $_POST['password'];
$confirm_password = $_POST['confirm_password'];
$sqli = "UPDATE detail SET password='$password',cnf_pass='$confirm_password' WHERE user_id=$id";
if (mysqli_query($con, $sqli)) {
echo 'record added';
} else {
echo 'error';
}
mysqli_close($con);
}
?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1" />
<title>Form 3 </title>
</head>
<body>
<div class="wrapperDiv">
<form action="process-complete.php" method="post" id="form">
<?php
if(!isset($_POST['submitBtn3'])) header('location: form-page2.php');
foreach($_POST as $key => $value) {
if($key != 'submitBtn3') {
?>
<input type="hidden" name="<?php echo $key; ?>" value="<?php echo $value; ?>" />
<?php
}
}
?>
<table width="500">
<tr>
<th colspan="3" scope="row"><h3>Form 3</h3></th>
</tr>
<tr>
<th scope="row"><div align="right">City</div></th>
<td><div align="center"><strong>:</strong></div></td>
<td><input type="text" name="city" value="" class="required" /></td>
</tr>
<tr>
<th scope="row"><div align="right">Street</div></th>
<td><div align="center"><strong>:</strong></div></td>
<td><input type="text" name="street" value="" class="required" /></td>
</tr>
<tr>
<th scope="row"><div align="right">District</div></th>
<td><div align="center"><strong>:</strong></div></td>
<td><input type="text" name="district" value="" class="required" /></td>
</tr>
<tr>
<th scope="row"> </th>
<td> </td>
<td><input type="submit" name="submitBtn2" id="submitBtn2" value="Save" /></td>
</tr>
</table>
</form>
</div>
</body>
</html>
這里$ id顯示為未定義變量。
終於，我找到了解決方案。 借助session變量，我可以在每個網頁上移動id 。
PHP-如何第一次插入表單數據，第二次更新相同數據
[英]PHP- how to insert form data first time and second time update same data
我如何從數據庫中獲取信息，就像我將數據與存儲的信息插入數據庫中以更新數據庫一樣
[英]How can I get the information form the database as the same form as I insert the data on it with the stored information in order to update it
我如何在不將那些數據記錄到數據庫的情況下將第一種格式的數據傳遞給第二種格式？
[英]how can i pass the data of the first form to the second one without recording those data on the database yet?
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