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我正在學習PHP,html和sql。我正在嘗試創建一個允許用戶更新現有客戶數據的表單。我想知道我的代碼出了什麼問題,因爲每次我嘗試對現有客戶進行更改時,這些更改都反映在數據集的第一個客戶上。例如,當我更改客戶ID#14的數據時,更改顯示在客戶ID#1中。客戶ID#14仍保留舊數據。這是我的代碼。有人可以給我一些關於如何解決它的提示嗎?我花了幾個小時,但我不知道我錯過了什麼。非常感謝!如何更新所選行的數據?
這是main.php文件中代碼的一部分。
<div>
<form method="post" action="updatecustomer.php">
<fieldset>
<legend>Update Existing Customer</legend>
<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>
<li>Customer ID:
<select name="customer_id">
<?php
if(!($stmt = $mysqli->prepare("SELECT customer_id, 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, $customer_id)){
echo "Bind failed: " . $mysqli->connect_errno . " " . $mysqli->connect_error;
}
while($stmt->fetch()){
echo '<option value=" '. $customer_id . ' "> ' . $customer_id . '</option>\n';
}
$stmt->close();
?>
</select>
</li>
</fieldset>
<input type="submit" name="update" value="Update Customer">
</div>
這裏是updatecustomer.php代碼:
if(!($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 . " " . $stmt->error;
}
if(!($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.";
}
爲什麼你在相同的查詢中兩次獲取customer_id? '從...選擇customer_id已經足夠了。 –
您的customer_id變量應該包含您想要修改的客戶的ID(在您的示例中爲14),但它似乎包含另一個(在您的案例中爲1)。一定要將正確的customer_id傳遞給您的查詢。 – eternay
@eternay:我對php和mysqli很陌生。你是否介意給我更多關於如何將正確的customer_id傳遞給我的查詢的細節?謝謝! – user2203774