Question

我有一个列名和列数据类型的数组，现在我希望使用这两个数组创建一个mysql表。到目前为止，这是我的代码：

<?php

//print_r($_GET);
$col_names=[]; //this will store column names received from user
$col_types=[];//this will store column data types selected by user


if(isset($_GET['col_num'])){

$table_name=$_GET['table_name'];
$n=$_GET['col_num'];


for($i=0;$i<$n;$i=$i+1){
    $index_names = "col".$i;
    $index_type = "type".$i;
    $col_names[$i] = $_GET[$index_names];
    $col_types[$i] = $_GET[$index_type];
}

}

$con=mysqli_connect('localhost','root');
if(!$con){
die("Error conncecting: ". mysqli_error($con));
}
else{
mysqli_select_db($con,'temp');

$query = "CREATE TABLE $table_name ( 
for($i=0; $i<$n ;$i=$i+1)
{
 echo "$col_names[$i]" . " " . "$col_types[$i]" . "(10)"  
}
);";
/*
  If suppose the col_names array contains : [Name,Age] and col_types   contains: [Varchar,Int] then i need these two attributes to be incorporated in my Create query and so  i have put them in a for loop.
*/

mysqli_query($query);
}
?>

现在我知道我写的“创建查询”出了问题，但是我无法弄清楚如何构建查询。另外，如果有多列，我应该如何放置逗号？

Answer 1

你做错了，使用这样的东西，

$query = "CREATE TABLE $table_name ( ";
for($i=0; $i<$n ;$i=$i+1)
{
  $query .= "$col_names[$i]" . " " . "$col_types[$i]" . "(10)"  ;
}
$query .= " ); ";
echo $query;//output and check your query
mysqli_query($query);

如何使用php动态创建mysql表？

1 个答案: