2 回答

TA貢獻(xiàn)1982條經(jīng)驗(yàn) 獲得超2個贊

TA貢獻(xiàn)1921條經(jīng)驗(yàn) 獲得超9個贊
1.password VARCHAT(20) NOT NULL這里,VARCHAAR打錯,應(yīng)該是password VARCHAR(20) NOT NULL,
2.mysqli_query($conn, $sql) 這里有問題,直接 mysql_query($sql);即可
下面為修改后的內(nèi)容
<?php
$host = "mysql.ict.swin.edu.au";
$user = "s7459394";
$pswd = "091290";
$dbnm = "s7459394_db";
$conn =@mysqli_connect($host,$user,$pswd);
if($conn === false)
{
die('Could not connect: ' . mysql_error());
}
mysql_select_db($dbnm);
$sql = "CREATE TABLE IF NOT EXISTS ssfriends(friend_id INT NOT NULL AUTO_INCREMENT PRIMARY KEY,friend_email VARCHAR(50) NOT NULL, password VARCHAR(20) NOT NULL, profile_name VARCHAR(30) NOT NULL, date_started DATE NOT NULL, num_of_friends INTEGER UNSIGNED )";
mysql_query($sql); //執(zhí)行SQL
echo "<p>Successfully created the table.</p>";
$sql = "CREATE TABLE IF NOT EXISTS myfriends( friend_id1 INT NOT NULL, friend_id2 INT NOT NULL )";
mysql_query($sql); //執(zhí)行SQL
echo "<p>Successfully created the table.</p>";
mysql_close();
?>
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