我做了一個非常簡單的代碼,以表格格式從數據庫檢索數據。但它不起作用,我不知道它有什麼問題。 以下是完整的PHP代碼我使用:簡單檢索數據庫中的數據mysql php
<?php
$link = mysql_connect('localhost','root', '');
if (!$link) {
die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
$db_selected = mysql_select_db('prelimdatabse', $link);
if (!$db_selected) {
die ('Can\'t use for: '. mysql_error());
}
$result = mysql_query("SELECT * FROM $registertable",$link);
echo "<table border='1'>
<tr>
<th>Username</th>
<th>Password</th>
<th>Firstname</th>
<th>Middlename</th>
<th>Lastname</th>
<th>Gender</th>
<th>Email</th>
<th>Month</th>
<th>Date</th>
<th>Year</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['uname'] . "</td>";
echo "<td>" . $row['pword'] . "</td>";
echo "<td>" . $row['fname'] . "</td>";
echo "<td>" . $row['mname'] . "</td>";
echo "<td>" . $row['lname'] . "</td>";
echo "<td>" . $row['gender'] . "</td>";
echo "<td>" . $row['email'] . "</td>";
echo "<td>" . $row['month'] . "</td>";
echo "<td>" . $row['date'] . "</td>";
echo "<td>" . $row['year'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($link);
?>
輸出:
Username Password Firstname Middlename Lastname Gender Email Month Date Year "; while($row = mysql_fetch_array($result)) { echo ""; echo "" . $row['uname'] . ""; echo "" . $row['pword'] . ""; echo "" . $row['fname'] . ""; echo "" . $row['mname'] . ""; echo "" . $row['lname'] . ""; echo "" . $row['gender'] . ""; echo "" . $row['email'] . ""; echo "" . $row['month'] . ""; echo "" . $row['date'] . ""; echo "" . $row['year'] . ""; echo ""; } echo ""; mysql_close($link); ?>
是的,這是代碼的結果......當它運行它只會顯示得那樣。
使用mysqli_query();和mysqli_fetch_array($結果)嘗試 –
我應該改變所有的MySQL到mysqli? – ChocoLover
請在'mysql_connect(SOMETHING)'和'mysql_query(SOMETHING)'後面添加'OR DIE(mysql_error())'並編輯您的帖子並創建錯誤。此外,它認爲**糟糕的做法**仍然使用過時的'mysql_FUNCTIONS'而不是'mysqli'。 –