I have a piece of code that should get the names of food items that are not ingredients from one table and then generate a query to create another table with columns matching the names of those items this is not generating a table:
function create_table(){
$query = "SELECT * FROM food";
$tablequery = "CREATE TABLE (id int(6) NOT NULL auto_increment,";
if($result=mysql_query($query)){
while($row = mysql_fetch_array($result)){
if($result['ingredient'] != 1){
$tablequery = $tablequery.$result['name']."varchar(30) NOT NULL,";
}
}
$tablequery = $tablequery."PRIMARY KEY (id),UNIQUE id (id),KEY id_2 (id))";
}
mysql_query($tablequery);
}
I did call the function so that is not the problem and somehow it worked once but with just the id.
Try this:
while($row = mysql_fetch_assoc($result)){
if($row['ingredient'] != 1){
$tablequery = $tablequery.$row['name']."varchar(30) NOT NULL,";
}
}
我认为您需要在代码中添加一行以删除表(如果该表已经存在):
DROP TABLE IF EXISTS `<your table>`;
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