How to create table using this below code

Question

1.00/5 (1 vote)

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Sir,
please tell how to create table using below code

What I have tried:

$table_name = "test";
$a ="r1,r2,r3,r4";
$abc = (explode(",",$a));
foreach($abc as $x)
{

echo $x;
$query = "CREATE TABLE $table_name".
'(id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
'.$x.' varchar(30))'

if($con->exec($query))
{
echo "Table MyGuests created successfully";
}
else
{
echo "try again";
}

}

after this the table should created in database but it is not.Why I am getting the solution.Please guide me

Posted 30-Nov-16 21:03pm

Member 12880505

Updated 1-Dec-16 0:57am

v6

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Comments

Peter Leow 1-Dec-16 6:41am

I can see that you have amended your code based on my Solution 1. But then, look carefully at your amended code, you are very careless by introducing new mistakes. I'm sure you have encountered error messages, look at the error messages and debug it yourself. Do not too quick to jump to ask question. You can't move too far this way.

2 solutions

Solution 2

$table_name = "test";
$a ="r1,r2,r3,r4";

i want to explode this
$products = (explode(",",$a));

after exploding i will get the answer as

foreach($a as $x)
{
echo $x;
}

$x through these variable i want to create column in table in database.

i wrote the query as follow

SQL

"CREATE TABLE $table_name".
    '(id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
    '.$x.' varchar(30))'

after this the table should be created in database but it is not.Why I am getting the solution.Please guide me

Posted 30-Nov-16 23:56pm

Member 12880505

Updated 1-Dec-16 1:05am

v2

Comments

Peter Leow 1-Dec-16 6:42am

This is not the solution. Use the improve question button instead.

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Peter Leow · Accepted Answer · 2016-11-30T23:02:00

Look into these three points:
1. What is

$a = r1,r2,r3,r4;

? Do you mean

$a = "r1,r2,r3,r4";

?
2. The $xyz is assigned the <p> tag:

$xyz = "<p>".$abc."</p>";

which got included into the field name.

SQL

"CREATE TABLE $table_name".
    			'(id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
			  	  '.$xyz.' varchar(30))'

I do not think that is what you want.
3. What is the purpose of $x? It did not seem to do anything useful.