Dynamic INSERT with full variables in PHP - php

Question

I would appreciate it if someone can guide me with the following problem, I need to do a INSERT into a MySQL database with variables, here is my code example:
$valor = "$valor.", ".$total;
$table = $dta.$year;
$insert = "ano, id, idas".$valor2.", total";
$value = $year.", ".$id.", ".$idas.$tit2.", ".$monto3;
$sql = "INSERT INTO `$table` (`$insert`) VALUES (`$value`)";
The general idea is that the amount of columns are not known until the program is run as it is all dynamic.
The following code is used to build the table as needed, which works fine:
while(($row = mysql_fetch_array($result)) !== FALSE)
{
$columna=$row['pn'];
$valor= $valor." pt".$columna." VARCHAR(50),";
$valor2= $valor2.", pt".$columna;
$gestion = '3';
$table = "sumag".$gestion.$year;
}
$valor= "ano VARCHAR(4), id VARCHAR(4), idas VARCHAR(13), ".$valor." "."total VARCHAR(10)";
For this reason the first code also has to be dynamic..
I really need help on this, I am not a programmer and I have built this from what I have learnt, so most probally my tecnique is pretty bad, if anyone can help it would be great..
Thank You

Answer 1

You have declared the column name as idas VARCHAR(13), in table but the insert element declared with some additional values idas".$valor2.". Check this one instance right now..

Answer 2

Here is the full code:
<?php
$gestion = 3;
$data = "SELECT * FROM progdesc where gestion = '$gestion'" or die("¡Error No existe ese departamento!");
$result = mysql_query($data) or die("¡Error No existe ese dep!");
while(($row = mysql_fetch_array($result)) !== FALSE)
{
$columna=$row['pn'];
$valor= $valor." pt".$columna." VARCHAR(50),";
$valor2= $valor2.", pt".$columna;
$gestion = '3';
$table = "sumag".$gestion.$year;
}
$valor= "ano VARCHAR(4), id VARCHAR(4), idas VARCHAR(13), ".$valor." "."total VARCHAR(10)";
#echo $table." - ".$valor;
mysql_query("CREATE TABLE `".$table."`(".$valor.")") or die(mysql_error());
//------------------------------------Insert data -----------------------------------------//
//-1--------Get Asig N°--------//
$data1 = "SELECT * FROM asigna" or die("¡Error No existe ese departamento!");
$result1 = mysql_query($data1) or die("¡Error1!");
while(($row = mysql_fetch_array($result1)) !== FALSE)
{
$id=$row['id'];
$idas=$row['idas'];
$sid=$row['sid'];
$tit2 = NULL;
$monto3 = 0;
//-2-------Get Gestion and DEPFINAL---------------------//
$data2 = "SELECT * FROM progdesc WHERE gestion = '$gestion' and ano = '$year' " or die("¡Error No existe ese departamento!");
$result2 = mysql_query($data2) or die("¡Error2!");
while(($row2 = mysql_fetch_array($result2)) !== FALSE)
{
$dep=$row2['departamento'];
$titulo=$row2['titulo'];
$depf = $dep."final";
$pn = $row2['pn'];
$sidb = $row2['sid'];
#$monto3 = 0;
#echo $pn;
//-3-------Get Program N° and create chain------------//
$data3 = "SELECT * FROM $depf WHERE ano = '$year' and idas = '$idas' and gestion = '$gestion'and programa = '$pn' " or die("¡Error No existe ese departamento!");
$result3 = mysql_query($data3) or die("¡Error3!");
while(($row3 = mysql_fetch_array($result3)) !== FALSE)
{
$monto=$row3['monto'];
if (!($idas2 == $idas and $monto == 0)){
$tit2 = $tit2.", ".$monto;
}elseif ($monto == 0){
$tit2 = $tit2.", 0";
} else {
$cero = 0;
$pn2 = 0;
If (!($monto == 0)){
for ($i = 0; $i < $pn2; $i++) {
$cero = $cero.", 0";
}
$tit2 = $tit2.", ".$cero.", ".$monto;
}
}
$idas2 = $idas;
$ln = $ln + 1;
$monto2 = $monto;
$monto3 = $monto3 + $monto;
}
}
$insert = "ano, id, idas".$valor2.", total";
$value = $year.", ".$id.", ".$idas.$tit2.", ".$monto3;
$sql = "INSERT INTO `$table` (`$insert`) VALUES (`$value`)";
myquery($sql); }

Answer 3

I just want to make clear that I am not a programmer and I know my code is not up to standard in any way, I was forced to complete this task just because I kinda know programming, this is why I am asking for help, if there is anther way to complete the task, great and thanks fr all have responded so far, I apreciate your patience.