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$uid = $_GET['id'];
echo $sql = "select * from notice where id=$uid";
echo $result = mysql_query($sql);
$row = mysql_fetch_array($result);
    echo $row['headline'];
    echo $row['description'];
    echo $row['image'];

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Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\willammary\staff-admin\edit_notice.php on line 5
4

1 回答 1

0

尝试这个,

$uid = $_GET['id'];
echo $sql = "select * from notice where id=".$uid;
echo $result = mysql_query($sql);
if($result === FALSE) {
    die(mysql_error()); // error handling
}
$row = mysql_fetch_array($result);
echo $row['headline'];
echo $row['description'];
echo $row['image'];

或者,

$uid = $_GET['id'];
echo $sql = "select * from notice where id=$uid";
echo $result = mysql_query($sql);
if($result) {
   $row = mysql_fetch_array($result);
   echo $row['headline'];
   echo $row['description'];
   echo $row['image'];
}

或者

$uid = $_GET['id'];
echo $sql = "select * from notice where id=$uid";
echo $result = mysql_query($sql) or die(mysql_error());
$row = mysql_fetch_array($result);
echo $row['headline'];
echo $row['description'];
echo $row['image'];

mysql 扩展也是从PHP 5.5.0deprecated开始阅读http://www.php.net/manual/en/intro.mysql.php

于 2013-08-22T11:02:34.127 回答