所以我很难得到一个使用mysql搜索命令的变量,然后在插入命令的相同脚本中使用它。我究竟做错了什么?获取mysql结果并在同一个脚本中的后续插入命令中使用它
<?php
$usto= $_GET["usto"];
$itena= "item";
$sql = 'SELECT sname FROM login';
$hostname_Database = "blocked";
$database_Database = "blocked";
$username_Database = "blocked";
$password_Database = "blocked";
$mysqli = new mysqli($hostname_Database, $username_Database, $password_Database, $database_Database);
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$result = $mysqli->query($sql);
if ($result) {
$row = $result->fetch_assoc();
$sql = "INSERT INTO pon(mis, take)
VALUES({$row['snake']}, '" . $usto . "')"; //Here, I am trying to use the result from the previous select statement for the variable
$result = $mysqli->query($sql);
if ($result) {
...etc.
}
}
?>
''SELECT sname FROM login WHERE sname ='$ itena'';这是不是会引发错误? – crush