在PHP中使用mysql_fetch_assoc时发出警告 [英] Warning when using mysql_fetch_assoc in PHP
问题描述
Possible Duplicate:
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result
当我运行php页面时,出现此错误,并且不知道出了什么问题,有人可以帮忙吗?如果有人需要更多信息,我将发布整个代码.
When I run my php page, I get this error and do not know what's wrong, can anyone help? If anyone needs more infomation, I'll post the whole code.
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in
H:\Program Files\EasyPHP 2.0b1\www\test\info.php on line 16
<?PHP
$user_name = "root";
$password = "";
$database = "addressbook";
$server = "127.0.0.1";
$db_handle = mysql_connect($server, $user_name, $password);
$db_found = mysql_select_db($database, $db_handle);
if ($db_found) {
$SQL = "SELECT * FROM tb_address_book";
$result = mysql_query($SQL);
while ($db_field = mysql_fetch_assoc($result)) {
print $db_field['ID'] . "<BR>";
print $db_field['First_Name'] . "<BR>";
print $db_field['Surname'] . "<BR>";
print $db_field['Address'] . "<BR>";
}
mysql_close($db_handle);
}
else {
print "Database NOT Found ";
mysql_close($db_handle);
}
?>
推荐答案
通常表示您的SQL错误.
It generally means that you've got an error in your SQL.
$sql = "SELECT * FROM myTable"; // table name only do not add tb
$result = mysql_query($sql);
var_dump($result); // bool(false)
很明显,false
不是MySQL资源,因此会出现该错误.
Obviously, false
is not a MySQL resource, hence you get that error.
编辑立即粘贴的代码:
在while
循环之前的行上,添加以下内容:
On the line before your while
loop, add this:
if (!$result) {
echo "Error. " . mysql_error();
} else {
while ( ... ) {
...
}
}
确保tb_address_book
表确实存在,并且您已正确连接到数据库.
Make sure that the tb_address_book
table actually exists and that you've connected to the DB properly.
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