為什么我的插入數據代碼不起作用？
[英]Why isn't my insert data code working?
問題描述
我正在嘗試將數據插入數據庫 mysql phpAdmin。
我的虛擬主機是 000webhost。
我連接mysql數據庫的代碼：
<?PHP
$mysql_host = "mysql2.000webhost.com";
$mysql_database = "*********";
$mysql_user = "********";
$mysql_password = "**********";
$dbcon = mysql_connect($mysql_host,$mysql_user,$mysql_password,$mysql_database);
if (!$dbcon) {
die('error connecting to database');
}
echo ('You have connected successfully');
?>
我的插入數據代碼：
<?PHP
if (isset($_POST['submitted'])) {
include('connect_mysql.php');
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$sqlinsert = "INSERT INTO people (firstname, lastname) VALUES ('$fname', '$lname')";
if (!mysql_query($dbcon, $sqlinsert)) {
die('error inserting new record');
} // end of nested if statement
$newrecord = "1 record added to the database";
}
?>
<html>
<head>
<title>Insert Data into DB</title>
</head>
<body>
<h1>Insert Data into DB</h1>
<form method="post" action="insert-data.php">
<input type="hidden" name="submitted" value="true" />
<fieldset>
<legend>New People</legend>
<label>First Name: <input type="text" name="fname" /></label>
<label>Last Name: <input type="text" name="lname" /></label>
</fieldset>
<br />
<input type="submit" value="add new person" />
</form>
<?PHP
echo $newrecord
?>
</body>
</html>
而不是讓我把它放入數據庫它把我帶到這個頁面http://error404.000webhost.com/ ？
4 個解決方案
解決方案1
1 2014-04-14 09:04:16
嘗試改變
<label>First Name: <input type="text name="fname" /></label>
<label>Last Name: <input type="text name="lname" /></label>
到
<label>First Name: <input type="text" name="fname" /></label>
<label>Last Name: <input type="text" name="lname" /></label>
並插入查詢和連接到
$dbcon = mysql_connect($mysql_host,$mysql_user,$mysql_password,$mysql_database);
$sqlinsert = "INSERT INTO people (firstname, lastname) VALUES ('$fname', '$lname')";
解決方案2
0 2014-04-14 09:03:03
首先更改以下代碼：
$dbcon = mysqli_connect($mysql_host,$mysql_user,$mysql_password,$mysql_database);
到 ：
$dbcon = mysql_connect($mysql_host,$mysql_user,$mysql_password,$mysql_database);
然后更改以下代碼：
$sqlinsert = "INSERT INTO people (firstname, lastname) VALUES ('fname', 'lname')";
到：
$sqlinsert = "INSERT INTO people (firstname, lastname) VALUES ('$fname', '$lname')";
解決方案3
0 2014-04-14 09:09:23
你能改變這條線嗎？
<fieldset>
<legend>New People</legend>
<label>First Name: <input type="text name="fname" /></label>
<label>Last Name: <input type="text name="lname" /></label>
</fieldset>
到
<fieldset>
<legend>New People</legend>
<label>First Name: <input type="text" name="fname" /></label>
<label>Last Name: <input type="text" name="lname" /></label>
</fieldset>
你設置了錯誤的屬性，所以它沒有得到值..
之后，您必須更新您的 sql 查詢：
$sqlinsert = "INSERT INTO people (firstname, lastname) VALUES ('$fname', '$lname')";
解決方案4
0 已采納 2014-04-17 04:58:56
它確實必須是 mysqli
我的問題是 - 必須在<form method="post" action="insert-data.php">上更改為 _
工作代碼：
<?PHP
if (isset($_POST['submitted'])) {
include('connect_mysql.php');
$fname = $_POST['fname'];
$lname = $_POST['lname'];
$sqlinsert = "INSERT INTO people (firstname, lastname) VALUES ('$fname', '$lname')";
if (!mysqli_query($dbcon, $sqlinsert)) {
die('error inserting new record');
}
$newrecord = "1 new record added to the database";
}
?>
<html>
<head>
<title>Insert Data into DB</title>
</head>
<body>
<h1>Insert Data into DB</h1>
<form method="post" action="insert_data.php">
<input type="hidden" name="submitted" value="true" />
<fieldset>
<legend>New People</legend>
<label>First Name: <input type="text" name="fname" /></label>
<label>Last Name: <input type="text" name="lname" /></label>
</fieldset>
<br />
<input type="submit" value="add new person" />
</form>
<?PHP
echo $newrecord
?>
</body>
</html>
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