如何使用php变量作为表名在mysql中插入记录

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();
?>

Answer 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')";

Answer 2

尝试使用“ INSERT INTO karthick。”。$ id。”（名称，罐头，地址，产品，发票，运输

Answer 3

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')";

Answer 4

$val = 'karthick'.$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')";

Answer 5

尝试

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

Answer 6

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."`";

Answer 7

您正在创建karthick.details表，而不是karthick.session_id