我创建此代码是为了在数据库中添加一些数据而不重复
但这是不可行的,我不能将这个php创建的数据添加到我的表中
怎么了
P.S:我没有错
附言:我在wamp服务器上做这个。
<?php
$con=mysqli_connect("localhost","root","","kosar");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
for ($i=0;$i<10;$i++) {
$n1 = rand(100000,999999);
$n2 = rand(100000,999999);
$pincode = (string)$n1;
echo $pincode;
echo nl2br("$pincode'n");
mysql_query ("INSERT IGNORE INTO kosar(ID, Code, Type, Used) VALUES('', '".$pincode."','1', '0')");
if(mysql_affected_rows() == 0)
{
// Duplicates found
}else{
// No duplicates were found
}
}
for ($i=0;$i<10;$i++) {
$n1 = rand(100000,999999);
$n2 = rand(100000,999999);
$pincode = (string)$n1;
echo $pincode;
echo nl2br("$pincode'n");
mysql_query("INSERT IGNORE INTO kosar(ID, Code, Type, Used) VALUES('', '".$pincode."','2', '0')");
if(mysql_affected_rows() == 0)
{
// Duplicates found
}else{
// No duplicates were found
}
}
for ($i=0;$i<10;$i++) {
$n1 = rand(100000,999999);
$n2 = rand(100000,999999);
$pincode = (string)$n1;
echo $pincode;
echo nl2br("$pincode'n");
mysql_query("INSERT IGNORE INTO kosar(ID, Code, Type, Used) VALUES('', '".$pincode."','3', '0')");
if(mysql_affected_rows() == 0)
{
// Duplicates found
}else{
// No duplicates were found
}
}
mysqli_close($con);
?>
我很惊讶这没有引发任何错误。正如Barmar所提到的,不能混合使用sqli
和sql
函数。
同样从逻辑上讲,我看不出您选择任何数据库。
尝试第一次
$con=mysqli_connect("localhost","my_user","my_password","my_db") or die('error message');
然后使用
mysqli_query($con,"your query here");
让我知道结果。