Insert data php mysql

Question

Here is my full code for my problem and I will explain it:

<form name="form1" action="" method="POST">
<table>
  <tr>
    <th>ORDER TYPE</th>
    <th>ITEM CODE</th>
    <th>ITEM NAME</th>
    <th>ACTION</th>
  </tr>
  <?php
    $sql = "SELECT * FROM product";
    $i = 1; //to generate an unique id and name attr
    while($data = mysql_fetch_array($sql)){
       $itemcode = $data['cars_kd'];
       echo "<tr> <td><select name='select'><option>--Select--</option><option value='0'>Distributor</option><option value='1'>Retail</option>
       <input id='id$i' name='name$i' /></select></td>
       <td>$data['cars_kd']<input type='hidden' id='id$i' name='name$i' value='$itemcode'/></td>
       <td>$data['cars_name']</td>
       <td><input type='checkbox' name='check'/><input type='hidden' id='id$i' name='name$i'/></td> </tr> $i++";
     }
     ?>
     <input type="submit" name="save" value="Save" />
   </table>

In my PHP code above, I use $i to generate an unique id and name attr. After all of the data shown I'll choose some of that data which I want it by checked the checkbox. In my code above I have three input tag. 1. function is to hold the value of the select when checkbox is check. 2. to display $data['cars_kd'] function is to show all of the data record. the value of the input is what I take to then I show well in the third input. 3. function is as parameter of user choice. So, data which I wan to insert is order type (taken from select option value) and item code (taken from input three). I use input because I think I must have a parameter to know what the user choosing.

I hope everyone here can understand my explanation. Hope any suggest :-). Thanks.

Answer 1

1. Give action in your form action.Right now its empty

2. You need to give the name to your all text boxes name[] not name$i that means:-

Change <input type="text" id="id$i" name="name$i" />$data['data']; to <input type="text" id="id$i" name="name[]" />$data['data'];

3. Now you will get all the names in $_POST['name'] array.

4. I assumed that your table structure is (id,car_name) and you want to save data like (1,Volvo)(2,Dukati) etc. Also i assume that id is auto-incremented and primary key. Now you need to do stuff like below:-

   $_POST['name'] = array('Volvo','Dukati','KTR','BMW');// now your POST array will look like this
   $data = $_POST['name'];
   $values = '';
   foreach($data as $dat){
      $values .= "(".$dat."),";
   }
   $values = trim($values,",");
   
   $sql = "INSERT INTO <table name> ('car_name') VALUES '".$values."'";
   echo $sql;
   

Output:- https://eval.in/510646

Note:- Its not the complete code, its just an example to show you how can you proceed. Thanks

Answer 2

Put to action="" path to script like this /php/script.php

AND add <? echo ;?> for print variable $i value

**PHP **

 <form action="/script.php" method="POST">
   <input type="text" id="id<? echo $i; ?>" name="name<? echo $i; ?>" /><? echo $data['data'];?>

  <input type="submit" name="submit" value="Save"/>

Answer 3

If I understand your question, you can try in this way:

foreach( $_POST as $key => $val )
{
    if( preg_match( '/^name(\d+)$/',$key,$matches ) )
    {
        $name = $val;
    }
}

the id number is catched in $matched[1], so - if you want use it as array index:

       $name[$matches[1]] = $val;

Edit:

If you known the exact number of name postfields, you can also try in another way.
Supposing total field number is 5:

for( $i=1; $i<6; $i++ )
{
    $name = $_POST["name$i"];
}

Answer 4

<?php 
include('conect_to_db.php');
?>
<form name="form1" action="/a.php" method="POST">  <? /* action= the path + filename of script witch catch post data and work with it*/ ?>
<? for($i=0;$i<3;$i++) { ?>
<input type="text" id="id<? echo $i; ?>" name="name[]" value="<? echo $data  ['data'];?>"\>
<? 
}
?>
<input type="submit" name="submit" value="Save"/>
</form>

<?
if(isset($_POST['name'])) {
$data =$_POST['name'];
foreach($data as $value) {
/*may be like this to see you your darling name1,name2,name3, but you can do simply*/
if(!isset($name1)) $name1=mysql_real_escape_string($value); else
if(!isset($name2)) $name2=mysql_real_escape_string($value); else
if(!isset($name3)) $name3=mysql_real_escape_string($value); else 
return;
}

$sql = "INSERT INTO table_namex (name1,name2,name3) VALUES (`$name1`,`$name2`,`$name3`)";
$res=mysql_query($sql);
echo 'result(1=good): '.$res.' \   query: '.$sql;

}
?>