2014-09-20 110 views
-1

当我运行的代码,它显示了这个错误。“mysqli_query()预计参数1是mysqli的,空给”mysqli_query()预计参数1是mysqli的,空给出:(

<?php 

function confirm_query($result_set) { 
if (!$reslt_set) { 
    die("database query failed"); 
} 
} 

function find_all_subjects() { 
global $connection; 
$query = "SELECT * "; 
$query .= "FROM subjects "; 
$query .= "WHERE visible=1 "; 
$query .= "ORDER BY position ASC "; 
$subject_set = mysqli_query($connection, $query); 
confirm_query($subject_set); 
return $subject_set; 
} 
?> 
+0

我db_connetion是mysqli_ :(...这是代码 2014-09-20 13:46:28

回答

1

如果是这样的你的完整文件,然后$connection将是空的(空),你不创建连接任何地方。

您可能创建连接在另一个文件,因此global $connection;但您需要包括此文件获得连接

0

您正在混合mysql_mysqli函数。这可能是你的问题。你不能在代码中混合使用mysqli和mysql函数。

0

试试这个。

$link = mysqli_connect("localhost", "my_user", "my_password", "world"); 

/* check connection */ 
if (mysqli_connect_errno()) { 
    printf("Connect failed: %s\n", mysqli_connect_error()); 
    exit(); 
} 

/* Create table doesn't return a resultset */ 
if (mysqli_query($link, "CREATE TEMPORARY TABLE myCity LIKE City") === TRUE) { 
    printf("Table myCity successfully created.\n"); 
} 

/* Select queries return a resultset */ 
if ($result = mysqli_query($link, "SELECT Name FROM City LIMIT 10")) { 
    printf("Select returned %d rows.\n", mysqli_num_rows($result)); 

    /* free result set */ 
    mysqli_free_result($result); 
} 

得到这个从php.net只有

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