PHP MySQLi INSERT 不工作，没有错误
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PHP MySQLi INSERT not working, no errors
提问by Canadian Luke
Different from this question, but similar in that I don't get an error when adding information to my database.
与此问题不同，但相似之处在于，将信息添加到我的数据库时我没有收到错误消息。
$sql = "INSERT INTO 'nlcc_ver1'.'tUsers' ('userID', 'userName', 'userPassword', 'userHash',
'user_first_name', 'user_last_name', 'user_corps', 'is_admin', 'is_trg', 'is_sup', 'is_co')
VALUES (NULL, '" . $userName . "', '" . $hash . "', '" . $salt . "', '" . $f_name . "', '" .
$l_name . "', '" . $corps . "', '" . $admin . "', '" . $trg . "', '" . $sup . "', '" . $co . "')";
$hostname_Database = "localhost";
$database_Database = "nlcc_ver1";
$username_Database = "root";
$password_Database = "";
$mysqli = new mysqli($hostname_Database, $username_Database, $password_Database, $database_Database);
if (mysqli_connect_errno()) {
printf("Connect failed: %s\n", mysqli_connect_error());
exit();
}
$result = $mysqli_query($mysqli, $sql);
echo "Query run. Inserted UserID " . mysqli_insert_id($mysqli) . "<br />";
Line breaks inserted to avoid sideways scrolling... It says on the web page that mysqli_insert_id($mysqli) is 0, and nothing is added to the table on my database. I do not see an error connecting to the database appearing, and MySQL is running on my server, and phpinfo() shows both the MySQL and MySQLI extension loaded. This is just a development machine, so don't worry about the security (i.e. no password). I have tried googling the problem, but am not finding too much. I don't know about object oriented PHP programming with ->, I am used to using _. Is this method still supported?
插入换行符以避免横向滚动...它在网页上说 mysqli_insert_id($mysqli) 为 0，并且没有向我的数据库的表中添加任何内容。我没有看到连接到数据库的错误出现，而且 MySQL 正在我的服务器上运行，并且 phpinfo() 显示加载了 MySQL 和 MySQLI 扩展。这只是一个开发机器，所以不要担心安全性（即没有密码）。我试过用谷歌搜索这个问题，但没有找到太多。我不知道使用->的面向对象的PHP编程，我习惯使用_。这个方法还支持吗？
回答by Lightness Races in Orbit
You've mixed procedural and object-oriented MySQLi styles. This has led to you trying to use the functions like mysqli_query($mysqli)instead of the member functions like $mysqli->query(). Your $mysqliis an object, not a resource handle.
您已经混合了面向过程和面向对象的 MySQLi 风格。这导致您尝试使用像这样的函数mysqli_query($mysqli)而不是像$mysqli->query(). 你$mysqli是一个对象，而不是一个资源句柄。
And, you're not performing any error checking on your query. If you were, you'd see that you have mistakenly used single quotes to delimit table and field names, not backticks.
而且，您没有对查询执行任何错误检查。如果是，您会发现您错误地使用单引号来分隔表名和字段名，而不是反引号。
$sql = "INSERT INTO `nlcc_ver1`.`tUsers`
(`userID`, `userName`, `userPassword`, `userHash`,
`user_first_name`, `user_last_name`, `user_corps`,
`is_admin`, `is_trg`, `is_sup`, `is_co`)
VALUES (NULL, '" . $userName . "', '" . $hash . "', '" . $salt . "', '" .
$f_name . "', '" . $l_name . "', '" . $corps . "', '" . $admin .
"', '" . $trg . "', '" . $sup . "', '" . $co . "')";
$hostname_Database = "localhost";
$database_Database = "nlcc_ver1";
$username_Database = "root";
$password_Database = "";
$mysqli = new mysqli($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) {
printf("%s\n", $mysqli->error);
exit();
}
echo "Query run. Inserted UserID " . $mysqli->insert_id . "<br />";
I strongly suggest using the manualas your reference. It's quite clear on how to use these functions when you're using either procedural or object-oriented style MySQLi.
我强烈建议使用手册作为您的参考。当您使用过程式或面向对象风格的 MySQLi 时，如何使用这些函数非常清楚。
回答by Sabeen Malik
$mysqli_query($mysqli, $sql);
should be
应该
mysqli_query($mysqli, $sql);
OR
或者
$mysqli->query($sql);
AND later on
后来
$mysqli->insert_id();
回答by Dr.Molle
Look at this:
看这个：
'nlcc_ver1'.'tUsers'
You have to use backticks here as quote character:
您必须在此处使用反引号作为引号字符：
`nlcc_ver1`.`tUsers`
But however(assuming that the $ in $mysqli_query is just a typo): You will not get errors for the query , unless you use mysqli_error() right after executing the query.
但是（假设 $mysqli_query 中的 $ 只是一个错字）：您不会收到查询错误，除非您在执行查询后立即使用 mysqli_error()。
回答by HimalayanCoder
SET AutoCommit = 1 before inserting
插入前设置 AutoCommit = 1
$mysqli->query('SET AUTOCOMMIT = 1');
