有人可以告诉我如何将 ID 传递给 PHP/MySQL 中的 UPDATE 查询吗?每次我在表单的下拉框中选择“ID #3”时,我的更新查询总是读取 ID #1。我相信问题出在下面的代码中。我已经坚持了2天。我无法说出我的代码中缺少什么。非常感谢您的帮助。
<div>
<form method="post" action="updatecustomer.php">
<fieldset>
<legend>Update Existing Customer</legend>
<li>Customer ID:
<select name="customer_id">
<?php
if(!($stmt = $mysqli->prepare("SELECT customer_id FROM customer"))){
echo "Prepare failed: " . $stmt->errno . " " . $stmt->error;
}
if(!$stmt->execute()){
echo "Execute failed: " . $mysqli->connect_errno . " " . $mysqli->connect_error;
}
if(!$stmt->bind_result($customer_id)){
echo "Bind failed: " . $mysqli->connect_errno . " " . $mysqli->connect_error;
}
while($stmt->fetch()){
echo '<option value=".$customer_id."> '.$customer_id.'</option>\n';
}
$stmt->close();
?>
</select>
</li>
<li>First Name: <input type="text" name="fName"> Last Name: <input type="text" name="lName"</li>
<li>Email Address: <input type="text" name="email"></li>
<li>Phone Number: <input type="text" name="phone_number"></li>
<li>Street Number: <input type="text" name="address_no"> Street Line 1: <input type="text" name="address_street1"></li>
<li>Street Line 2 (Apt or Unit Number): <input type="text" name="address_street2"></li>
<li>City: <input type="text" name="address_city"> State: <input type="text" name="address_state"> Zip: <input type="text" name="address_zip"> </li>
</fieldset>
<input type="submit" name="update" value="Update Customer">
</div>
这是updatecustomer.php
文件:
<?php
//Turn on error reporting
ini_set('display_errors', 'On');
if(!($stmt = $mysqli->prepare("UPDATE customer SET fName=?, lName=?, email=?, phone_number=?, address_no=?, address_street1=?,
address_street2=?, address_city=?, address_state=?, address_zip=? WHERE customer_id=?"))){
echo "Prepare failed: " . $stmt->errno . " " . $stmt->error;
}
echo $_POST['customer_id'];
if(!($stmt->bind_param("sssiissssii",$_POST['fName'],$_POST['lName'],$_POST['email'],$_POST['phone_number'], $_POST['address_no'],
$_POST['address_street1'],$_POST['address_street2'],$_POST['address_city'],$_POST['address_state'], $_POST['address_zip'], $_POST['customer_id']))){
echo "Bind failed: " . $stmt->errno . " " . $stmt->error;
}
if(!$stmt->execute()){
echo "Execute failed: " . $stmt->errno . " " . $stmt->error;
} else {
echo "Updated " . $stmt->affected_rows . " rows to customer.";
}
$stmt->close();
?>