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Im正在努力處理以下代碼。我的最終目標是創建一個簡短的div,在我的數據庫中顯示錶格中的信息。但它不工作。我是否也必須連接到特定的數據庫?從數據庫中選擇語句
<html>
<head>
</head>
<body>
<?php
$dbName = 'localhost';
$userName = 'root';
$passWord = 'mysql';
$conn = mysql_connect($dbName, $userName, $passWord);
// Check connection
if (!$conn)
{
die("Connection failed: " . $conn->connect_error);
}
echo "Connected successfully";
?>
<div id="displayAuthor">
<?php
$sql_statement = "
SELECT ssn, lastname, firstname
FROM author
ORDER
BY lastname, firstname
";
$result = mysql_query($sql_statement);
$outputDisplay = "";
if(!$result)
{
$outputDisplay .= "Error";
} else
{
$outputDisplay = "<h3> Table author data </h3>";
$outputDisplay .= "<tr><th>SSN</th> <th>Last name </th> <th> First name</th> </tr>";
$numberResults = mysql_num_rows($result);
for ($i=0; $i<$numberResults; $i++)
{
//Dit is een counter van hoeveel rijen het uiteindelijk was.
$row = mysql_fetch_array($result);
$ssn = $row['ssn'];
$lastname = $row['lastname'];
$firstname = $row['firstname'];
$outputDisplay .="<td>".$ssn."</td>";
$outputDisplay .="<td>".$lastname."</td>";
$outputDisplay .="<td>".$firstname."</td>";
$outputDisplay .= "</tr>";
}
$outputDisplay .="</table>";
}
print $outputDisplay;
?>
</div>
</body>
</html>
需要使用mysql_select_db來選擇你的數據庫。但不建議使用mysql_ *函數。改用pdo或mysqli_ *函數。 – fortune 2014-12-02 11:10:38
扔掉這個並使用[PDO](http://php.net/pdo)或[mysqli](http://php.net/mysqli);然後再問一次。 – 2014-12-02 11:10:40
您可以連接到特定數據庫,也可以在指定表格時指定數據庫,即查詢中的FROM FROM MyDatabase.MyTable – Eilidh 2014-12-02 11:11:06