我有一个php数组:
$cont_array = Array("613:m-ent:id=one","930:m-lk:id=one;x=180;y=79;which=1","1080:m-lev:id=one;");
我想把它插入一个mysql表,表名作为变量,比如 $table_name = "user1"
.
另外,我想用第一个 ":"
。在执行php代码之后,我得到的只是一个空表。
我需要帮助。这是我的密码:
<?php
$connection = mysqli_connect('localhost','root','','prueba');
$cont_array = Array("613:m-ent:id=one","930:m-clk:id=one;x=180;y=79;which=1","1080:m-lev:id=one;");
$table_name = "user1";
$sql_1 = "DROP TABLE IF EXISTS `".$table_name."`;";
$sql_1 .= "CREATE TABLE `".$table_name."` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`usercod` varchar(20) NOT NULL DEFAULT '',
`pid` varchar(1000) NOT NULL DEFAULT '',
`name` varchar(1000) NOT NULL DEFAULT '',
`time` varchar(1000) NOT NULL DEFAULT '',
`all_act` varchar(1000) NOT NULL DEFAULT '',
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=1 DEFAULT CHARSET=latin1;";
mysqli_multi_query($connection,$sql_1);
foreach ($cont_array as $row){
$break_e = explode(':', $row, 2);
$sql="INSERT INTO`".$table_name."` (usercode,pid,name,time,all_act) VALUES ("user","pid_value","user_name",'$break_e [0]','$break_e [1]');";
mysqli_query($connection,$sql);
}
?>
这就是我得到的:
这就是我想要的:
1条答案
按热度按时间lo8azlld1#
您的sql语句下面似乎有问题检查。