我正在尝试从MySQL表中选择数据,但是我收到了这样的警告:
注意:未定义的变量:导致第40行的C:\ wamp \ www \ widget_corp \ data_base.php
和这一个:
警告:mysqli_fetch_row()要求参数1为mysqli_result,在第40行的C:\ wamp \ www \ widget_corp \ data_base.php中给出null
这是我的代码:
$connection = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);
// Test if connection occured
if (mysqli_connect_errno()){
die ("Database connection failed: " . mysqli_connect_errno() . " (" .
mysqli_connect_errno() . ")" );
}
// Perform database query
$query = "SELECT * FROM subjects " ;
$resut = mysqli_query($connection, $query);
// Test if there was a query error
if (!$resut){
die ("Database connection failed!");
}
// Use return data if any
while ($row = mysqli_fetch_row($result)){
//output data from each row
var_dump($row);
echo "<hr />";
}
一切正常,直到while()
。任何建议/解决方案?提前谢谢!
答案 0 :(得分:1)
更加注意,你的语法错误:
$resut = mysqli_query($connection, $query);
将其更改为:
$result = mysqli_query($connection, $query);
并改为:
if (!$result){
die ("Database connection failed!");
}