执行php mysqli insert into table时出错

svujldwt 于 2021-06-17 发布在 Mysql

关注(0)|答案(1)|浏览(354)

我有一个表列

log_id

作为主键自动递增。
我是否需要在insert into语句中指定它？
尝试将值插入数据库时出现下一个错误：

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '(log_session,log_ip,log_vrijeme,log_model) my_log VALUES ('270043ae1526e4' at line 1 INSERT INTO (log_session,log_ip,log_vrijeme,log_model) my_log VALUES ('270043ae1526e44967889b4382ff69fd','93.142.54.135','2018-11-23 14:06:15','1402')

我的代码是：

<?php
$_SESSION['compare_hash'] = "270043ae1526e44967889b4382ff69fd";

$log_session = $_SESSION['compare_hash'];
$log_ip = $_SERVER['REMOTE_ADDR'];
$log_vrijeme = date("Y-m-d H:i:s");
$log_model = "1402";

$con = mysqli_connect("localhost","user","password","databse");
$sql = "
    INSERT INTO (log_session,log_ip,log_vrijeme,log_model) my_log
    VALUES ('$log_session','$log_ip','$log_vrijeme','$log_model')
";
$rez = mysqli_query($con, $sql) or die(mysqli_error($con)."<br>$sql");
mysqli_close($con);
?>

表结构：

Field           Type       Null    Key  Default     Extra   
log_id          int(11)     NO     PRI  NULL     auto_increment
log_session     varchar(42) NO     MUL  NULL    
log_ip          varchar(15) NO          NULL    
log_vrijeme     datetime    NO     MUL  NULL    
log_model       int(11)     NO     MUL  NULL

谢谢你的帮助。

mysql php mysqli error-handling

来源：https://stackoverflow.com/questions/53448227/error-while-doing-php-mysqli-insert-into-table

1条答案

按热度按时间

hi3rlvi21#

你应该把 my_log 刚好在…之后 INSERT INTO :

<?php
$_SESSION['compare_hash'] = "270043ae1526e44967889b4382ff69fd";

$log_session = $_SESSION['compare_hash'];
$log_ip = $_SERVER['REMOTE_ADDR'];
$log_vrijeme = date("Y-m-d H:i:s");
$log_model = "1402";

$con = mysqli_connect("localhost","user","password","databse");
$sql = "
    INSERT INTO my_log (log_session,log_ip,log_vrijeme,log_model)
    VALUES ('$log_session','$log_ip','$log_vrijeme','$log_model')
";
$rez = mysqli_query($con, $sql) or die(mysqli_error($con)."<br>$sql");
mysqli_close($con);
?>

但是，您的代码有几个缺点，稍后我将描述它们。
1不要将未转义的变量传递到查询中
除非你别无选择。使用 mysqli_real_escape_string($con, $your_var) 防止sql注入和使用pdo会使它变得更好。
2在以后的编辑过程中，使用herdeoc语法确保查询的一致性
这样即使你的某个人 " 在查询中，它仍将按预期工作。
三。如果下面没有任何内容，请省略php文件末尾的结束标记
这将防止意外错误的发生。

<?php
$_SESSION['compare_hash'] = "270043ae1526e44967889b4382ff69fd";
$con = mysqli_connect("localhost","user","password","databse");

$log_session = mysqli_real_escape_string($con, $_SESSION['compare_hash']);
$log_ip = mysqli_real_escape_string($con, $_SERVER['REMOTE_ADDR']);
$log_vrijeme = mysqli_real_escape_string($con, date("Y-m-d H:i:s"));
$log_model = mysqli_real_escape_string($con, "1402");

$sql = <<<SQL
    INSERT INTO my_log (log_session,log_ip,log_vrijeme,log_model)
    VALUES ('${log_session}','${log_ip}','${log_vrijeme}','${log_model}')
SQL;

$rez = mysqli_query($con, $sql) or die(mysqli_error($con)."<br>$sql");
mysqli_close($con);

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执行php mysqli insert into table时出错

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