[英]Fetching data from database using PHP
我試圖從elective_mgmt數據庫的優先級表中獲取數據。源代碼如下:
<?php
$connect = mysql_connect("localhost","root","");
mysql_select_db("elective_mgmt",$connect);
$result = mysql_query($con,"SELECT * FROM priority");
echo "<table border='1'>
`<tr>
<th>Name</th>
<th>Roll</th>
<th>Email</th>
<th>Priorityone</th>
<th>Prioritytwo</th>
<th>Prioritythree</th>
</tr>";
while($row = mysql_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['Name'] . "</td>";
echo "<td>" . $row['Roll'] . "</td>";
echo "<td>" . $row['Email']. "</td>";
echo "<td>" . $row['Priorityone']."</td>";
echo "<td" . $row['Prioritytwo']."</td>";
echo "<td" . $row['Prioritythree']."</td>";
echo "</tr>";
}
echo "</table>";
mysql_close($con);
?>
?>
當我運行它時,它顯示如下:
Warning: mysql_query() expects parameter 2 to be resource, string given in C:\xampp\htdocs\Elective_management\admin_view.php on line 5
Warning: mysql_fetch_array() expects parameter 1 to be resource, null given in C:\xampp\htdocs\Elective_management\admin_view.php on line 15
Name Roll Email Priorityone Prioritytwo Prioritythree
Warning: mysql_close() expects parameter 1 to be resource, null given in C:\xampp\htdocs\Elective_management\admin_view.php on line 28
?>
我什么都不知道 請幫我。
您對mysql_query
參數順序不正確。 首先查詢,然后是連接。
mysql_query("SELECT * FROM priority", $connect);
您的連接錯誤。 應該看起來像這樣
$result = mysql_query("SELECT * FROM priority",$connect );
因為您剛剛連接,所以您不需要連接變量。 您應該可以輸入
$result = mysql_query("SELECT * FROM priority");
並使其工作正常
mysql_close($con);
在這里,您沒有$ con變量,因此它為空,為什么會顯示此錯誤
Warning: mysql_close() expects parameter 1 to be resource, null given in C:\xampp\htdocs\Elective_management\admin_view.php on line 28
解決方法:更改
mysql_close($con);
至
mysql_close($connect);
$row = mysql_fetch_array($result)`
在這里,您將其指向$result
,在$result
您有$con = Null
因此顯示此錯誤
Warning: mysql_fetch_array() expects parameter 1 to be resource, null given in C:\xampp\htdocs\Elective_management\admin_view.php on line 15
解決:
當您修復第三步時將被修復
$result = mysql_query($con,"SELECT * FROM priority");
再次在這里,您有第二個參數作為字符串,它不應該是字符串,因此會出現此錯誤
Warning: mysql_query() expects parameter 2 to be resource, string given in C:\xampp\htdocs\Elective_management\admin_view.php on line 5
解決:
解決方法:更改
mysql_query($con,"SELECT * FROM priority");
至
mysql_query("SELECT * FROM priority",$connect)
PS:如果您只是開始在此項目中編碼
請考慮將語法從 MySql_ * 更改 為 PDO語法
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.