<?php
$q="localhost";
$w="root";
$p="19981111";
if(!mysqli_connect($q,$w,$p)){
die("fail");
}else{
echo"success";
}
if(mysqli_query("CREATE DATABASE database_name",mysqli_connect($q,$w,$p)
))
{echo"success";
}else{
echo"fail";}
mysqli_close(mysqli_connect($q,$w,$p));
?>为什么会出现Warning: mysqli_query() expects parameter 1 to be mysqli, string given in C:\wamp\www\789456123.php on line 12
$q = "localhost";
$w = "root";
$p = "19981111";
$link = mysqli_connect ( $q, $w, $p );
if (! $link) {
die ( "fail" );
} else {
echo "success";
}
if (mysqli_query ($link , "CREATE DATABASE database_name" )) {
echo "success";
} else {
echo "fail";
}
$w = "root";
$p = "19981111";
$link = mysqli_connect ( $q, $w, $p );
if (! $link) {
die ( "fail" );
} else {
echo "success";
}
if (mysqli_query ($link , "CREATE DATABASE database_name" )) {
echo "success";
} else {
echo "fail";
}
mysqli_query 你传的参数反了
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第1个回答 2015-06-12
mixed mysqli_query ( mysqli $link , string $query [, int $resultmode = MYSQLI_STORE_RESULT ] )
sql语句与数据库连接句柄参数位置搞反了！！
sql语句与数据库连接句柄参数位置搞反了！！
