使用php变量将值插入mysql表

Question

I am trying to use a variable to insert into multiple tables. When I hard code the specific table name it runs properly, when I use a variable I get a QUERY FAILEDSQLSTATE[42000]: Syntax error or access violation: 1064 error. dbname is the variable. I am using a for loop to change the name of the table. For example table 1 is budget1000, then budget 2000 etc. Here is my code

$sql='INSERT INTO ".$dbName." VALUES(:id,:category,:subCategory,
:amount, :today,:description,   :year)';


try{
$st= $conn->prepare($sql);
$st->bindValue(":id", $id, PDO::PARAM_INT);
$st->bindValue(":category", $category, PDO::PARAM_INT);
$st->bindValue(":subCategory", $subCategory, PDO::PARAM_INT);
$st->bindValue(":amount", $amount, PDO::PARAM_INT);
$st->bindValue(":today", $today, PDO::PARAM_STR);
$st->bindValue(":description", $description, PDO::PARAM_STR);
$st->bindValue(":year", $year, PDO::PARAM_INT); 
$st->execute();
}catch(PDOException $e ){
echo "QUERY FAILED" . $e->getMessage();

}

Answer 1

It looks like there's a quote mismatch, you start off with single quotes but then switch to double quotes when you concatenate the DB name into your string. Try replacing the single quotes at the beginning and end of your $sql string with double quotes and remove the periods around $dbname, or use single quotes all the way through.

Answer 2

试试这个：

$sql='INSERT INTO '.$dbName.' VALUES(:id,:category,:subCategory,:amount, :today,:description,   :year)';