[英]PHP, get data from the database
我想用 PHP 從數據庫中檢索數據並將其顯示在網站上。
此代碼無法正常工作。 我想在我的數據庫中顯示所有雪佛蘭汽車。
<?php
$db = mysqli_connect("localhost","myusername",
"mypassword","database");
if (mysqli_connect_errno()) {
echo("Could not connect" .
mysqli_connect_error($db) . "</p>");
exit(" ");
}
$result = mysqli_query($query);
if(!$result){
echo "<p> Could not retrieve at this time, please come back soon</p>" .
mysqli_error($dv);
}
$data = mysql_query("SELECT * FROM cars where carType = 'Chevy' AND active = 1")
or die(mysql_error());
echo"<table border cellpadding=3>";
while($row= mysql_fetch_array( $data ))
{
echo"<tr>";
echo"<th>Name:</th> <td>".$row['name'] . "</td> ";
echo"<th>ImagePath:</th> <td>".$row['imagePath'] . " </td></tr>";
echo"<th>Description:</th> <td>".$row['description'] . "</td> ";
echo"<th>Price:</th> <td>".$row['Price'] . " </td></tr>";
}
echo"</table>";
?>
如何使用 PHP 從數據庫中獲取數據?
你不是在查詢數據庫,所以它不會給你結果
這是它的工作原理
1)通過mysql_connect()
連接數據庫
mysql_connect("localhost", "username", "password") or die(mysql_error());
2)比選擇像mysql_select_db()
這樣的數據庫
mysql_select_db("Database_Name") or die(mysql_error());
3) 你需要使用mysql_query()
喜歡
$query = "SELECT * FROM cars where carType = 'chevy' AND active = 1";
$result =mysql_query($query); //you can also use here or die(mysql_error());
看看有沒有錯誤
4) 而不是mysql_fetch_array()
if($result){
while($row= mysql_fetch_array( $result )) {
//result
}
}
所以試試
$data = mysql_query("SELECT * FROM cars where carType = 'chevy' AND active = 1") or die(mysql_error());
echo"<table border cellpadding=3>";
while($row= mysql_fetch_array( $data ))
{
echo"<tr>";
echo"<th>Name:</th> <td>".$row['name'] . "</td> ";
echo"<th>ImagePath:</th> <td>".$row['imagePath'] . " </td></tr>";
echo"<th>Description:</th> <td>".$row['description'] . "</td> ";
echo"<th>Price:</th> <td>".$row['Price'] . " </td></tr>";
}
echo"</table>";
?>
Mysql_*
函數已棄用,因此請改用PDO
或MySQLi
。
我建議 PDO 它更容易和更容易閱讀你可以在這里學習MySQL 開發人員的 PDO 教程也檢查PDO 初學者(為什么?以及如何?)
<?php
// Connects to your Database
mysql_connect("hostname", "username", "password") or die(mysql_error());
mysql_select_db("Database_Name") or die(mysql_error());
$data = mysql_query("SELECT * FROM cars where cars.carType = 'chevy' AND cars.active = 1")
or die(mysql_error());
Print "<table border cellpadding=3>";
while($row= mysql_fetch_array( $data ))
{
Print "<tr>";
Print "<th>Name:</th> <td>".$row['name'] . "</td> ";
Print "<th>ImagePath:</th> <td>".$row['imagePath'] . " </td></tr>";
Print "<th>Description:</th> <td>".$row['description'] . "</td> ";
Print "<th>Price:</th> <td>".$row['Price'] . " </td></tr>";
}
Print "</table>";
?>
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