How to use php variables as table name to insert records in mysql

Question

$id = $_SESSION['id'];

i have the table name stored in the $id variable.

when i use the variable name with sql query it doesn't work

$sql = "INSERT INTO karthick.$id (name, tin, address, product, invoice, transport, cutting, amount, vat) VALUES ('$name', '$tin', '$address', '$product', '$invoice', '$transport', '$cutting', '$amount', '$vat')";

when i replace the karthick.$id as karthick.ford it works fine. but i want to use the variable stored in $id as my table name. how do i do it.

Edit--------------------------- my php code

<?php
session_start();
$id = $_SESSION['id'];
require 'database.php';
/*$sql = "CREATE TABLE karthick.details (id INT AUTO_INCREMENT PRIMARY KEY, name VARCHAR(50) NOT NULL)";
if($conn->query($sql)===TRUE){
    echo "table created";
}else{
    echo "table not created";
}*/

$name = mysqli_real_escape_string($conn, $_POST["cname"]);
$tin = mysqli_real_escape_string($conn, $_POST["tin"]);
$address = mysqli_real_escape_string($conn, $_POST["address"]);
$product = mysqli_real_escape_string($conn, $_POST["product"]);
$ddate = mysqli_real_escape_string($conn, $_POST["date"]);
$invoice = mysqli_real_escape_string($conn, $_POST["invoice"]);
$transport = mysqli_real_escape_string($conn, $_POST["transport"]);
$cutting = mysqli_real_escape_string($conn, $_POST["date"]);
$amount = mysqli_real_escape_string($conn, $_POST["amount"]);
$vat = mysqli_real_escape_string($conn, $_POST["vat"]);
$val = 'karthick'.$id;
echo $val;
$sql = "INSERT INTO $val (name, tin, address, product, invoice, transport, cutting, amount, vat) VALUES ('$name', '$tin', '$address', '$product', '$invoice', '$transport', '$cutting', '$amount', '$vat')";
if($conn->query($sql)===TRUE){
    echo "record inserted";
}else{
    echo "not inserted".$conn->error;
}
$conn->close();
?>

"INSERT INTO karthick".$id."

The Codesee
– The Codesee

2016-07-21 10:52:05 +00:00
Commented Jul 21, 2016 at 10:52 — The Codesee
– The Codesee, Commented Jul 21, 2016 at 10:52
$value = 'karthick'.$id; $sql = "INSERT INTO $value "

JYoThI
– JYoThI

2016-07-21 10:52:32 +00:00
Commented Jul 21, 2016 at 10:52 — JYoThI
– JYoThI, Commented Jul 21, 2016 at 10:52
Have you start session in your page??

Saty
– Saty

2016-07-21 10:53:44 +00:00
Commented Jul 21, 2016 at 10:53 — Saty
– Saty, Commented Jul 21, 2016 at 10:53
yes i stared my session @Saty

pkarthicbz
– pkarthicbz

2016-07-21 10:59:32 +00:00
Commented Jul 21, 2016 at 10:59 — pkarthicbz
– pkarthicbz, Commented Jul 21, 2016 at 10:59
mysqli_select_db($conn,"database_name");

JYoThI
– JYoThI

2016-07-21 11:14:37 +00:00
Commented Jul 21, 2016 at 11:14 — JYoThI
– JYoThI, Commented Jul 21, 2016 at 11:14

Rajkumar R · Accepted Answer · 2016-07-21 10:52:14Z

1

Try this one.

$val = 'karthick'.$id;
$sql = "INSERT INTO $val (name, tin, address, product, invoice, transport, cutting, amount, vat) VALUES ('$name', '$tin', '$address', '$product', '$invoice', '$transport', '$cutting', '$amount', '$vat')";

answered Jul 21, 2016 at 10:52

Rajkumar R

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1 Comment

pkarthicbz Over a year ago

it returns no database selected

Lynn Rey · Accepted Answer · 2016-07-21 10:53:47Z

0

Try using "INSERT INTO karthick.".$id." (name, tin, address, product, invoice, transport

answered Jul 21, 2016 at 10:53

Lynn Rey

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Comments

Rahul Patel · Accepted Answer · 2016-07-21 10:55:18Z

0

Place the $id in braces{} like karthick.{$id}. The query should be like below.

$sql = "INSERT INTO karthick.{$id} (name, tin, address, product, invoice, transport, cutting, amount, vat) VALUES ('$name', '$tin', '$address', '$product', '$invoice', '$transport', '$cutting', '$amount', '$vat')";

answered Jul 21, 2016 at 10:55

Rahul Patel

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Comments

JYoThI · Accepted Answer · 2016-07-21 10:55:32Z

0

$val = 'karthick'.$id;
$val = str_replace(" ","",$val);
 $sql = "INSERT INTO $val (name, tin, address, product, invoice, transport, cutting, amount, vat) VALUES ('$name', '$tin', '$address', '$product', '$invoice', '$transport', '$cutting', '$amount', '$vat')";

answered Jul 21, 2016 at 10:55

JYoThI

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3 Comments

JYoThI Over a year ago

mysqli_select_db($conn,"database_name");

pkarthicbz Over a year ago

yes i used mysqli_select_db($conn,"database_name"); it returns the same error

pkarthicbz Over a year ago

this is the error showing

not insertedYou have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ''karthick'.ford (name, tin, address, product, invoice, transport, cutting, a' at line 1

Saurabh · Accepted Answer · 2016-07-21 10:56:08Z

0

Try

$sql = "INSERT INTO karthick.{$id} (name, tin, address, product, invoice, transport, cutting, amount, vat) VALUES ('{$name}', '{$tin}', '{$address}', '{$product}', '{$invoice}', '{$transport}', '{$cutting}', '{$amount}', '{$vat}')";

answered Jul 21, 2016 at 10:56

Saurabh

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Comments

Lynn Rey · Accepted Answer · 2016-07-21 11:18:56Z

0

You are creating karthick.details table instead of karthick.session_id

answered Jul 21, 2016 at 11:18

Lynn Rey

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Comments

Robin Panta · Accepted Answer · 2016-07-21 13:43:29Z

0

First method :

Try putting the PHP variables inside curly braces {}

Like:

$sql = "INSERT INTO karthick.{$id} (name, tin, address, product, invoice, transport, cutting, amount, vat) VALUES ('{$name}', '{$tin}', '{$address}', '{$product}', '{$invoice}', '{$transport}', '{$cutting}', '{$amount}', '{$vat}')";

Second method :

Use PHP variables outside quotes

Like:

$sql = " INSERT INTO karthick.".$id." (name, tin, address .....

Update

try using `` to enclose your table like,

$val = "`karthick`.`".$id."`";

edited Jul 21, 2016 at 13:43

answered Jul 21, 2016 at 11:08

Robin Panta

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2 Comments

pkarthicbz Over a year ago

i tried your first, second and all answers provided here but it returns the same error.

not insertedYou have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near ' (name, tin, address, product, invoice, transport, cutting, amount, vat) VAL' at line 1

Robin Panta Over a year ago

try using `` to enclose your table like, $val = "karthick.".$id."";

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