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.

As said before, consider changing your input names to name[]. name$i wouldn't work since PHP only processes code within <?php ?> tags and name$i clearly isn't. You can get the values of the input array on $_POST by using the name of the input array as the first key and the index of the specific input control as the second (i.e. $_POST["name"][0]). Reference. — Revenant
– Revenant, Commented Feb 1, 2016 at 8:00
I said thanks to everybody here who have been given an answer :-) — Reza M
– Reza M, Commented Feb 2, 2016 at 9:01

Death-is-the-real-truth · Accepted Answer · 2016-02-01 07:24:34Z

3

Give action in your form action.Right now its empty
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'];

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

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

edited Feb 1, 2016 at 7:24

answered Jan 31, 2016 at 9:35

Death-is-the-real-truth

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5 Comments

Reza M Over a year ago

Is it same $_POST['name'] = array('Volvo','Dukati','KTR','BMW'); with $_POST['name'] = array($data['data']); ? Because I get Volvo, BMW, etc from my table and show it with while(mysql_fetch_array).

Michael Over a year ago

In example inserted value is (Volvo),(Dukati),(KTR),(BMW), but why you need array? You store data in DB like string?! Show your table structure?! I think you have : id=1;car=volvo;id=2;car=bmw; ... And you not need to insert in same row 2 car name?

Michael Over a year ago

If your table scructure is : id=1;car1=bmw;car2=zaporojets... you insert need to by insert into table (id,car1,car2) Values ('','bmw','zaporojets')

Reza M Over a year ago

I want to insert that in one field. Okay I'll update my code. Wait a minute.

Reza M Over a year ago

I said thanks to everybody here who have been given an answer :-)

Michael · Accepted Answer · 2016-01-31 09:31:21Z

1

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"/>

edited Jan 31, 2016 at 9:31

answered Jan 31, 2016 at 9:25

Michael

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2 Comments

Reza M Over a year ago

I think this :<input type="text" id="id<? echo $i; ?>" name="name<? echo $i; ?>" /><? echo $data['data'];?>. is same with mine: <input type="text" id="id$i" name="name$i" />$data['data']; because I make this tag in php array. But my problem with that such code is in my PHP code can't run it as $_POST["name$i"] even I've put it in array like @fusion3k suggest. Hope another suggest :-)

Michael Over a year ago

The example of @A-2-A is good. If you don't know how to use it close this site and take a book about PHP in hand and read it. you dont need $_POST["name$i"] use $_POST["name"] and fet all values name[] from form in one array and work with foreach to get all values

fusion3k · Accepted Answer · 2016-01-31 09:33:26Z

1

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"];
}

edited Jan 31, 2016 at 9:33

answered Jan 31, 2016 at 9:25

fusion3k

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3 Comments

Reza M Over a year ago

I've already try this : for( $i=1; $i<6; $i++ ) { $name = $_POST["name$i"]; } But it's not works @fusion3k

fusion3k Over a year ago

@RezaM I have tried and for me it works fine. BTW, one of the answers works or you are even now w/out a solution?

Reza M Over a year ago

I said thanks to everybody here who have been given an answer :-)

Michael · Accepted Answer · 2016-02-01 06:20:39Z

1

<?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;

}
?>

answered Feb 1, 2016 at 6:20

Michael

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2 Comments

Michael Over a year ago

+10 to @A-2-A because I used some code from he! This Is it same $_POST['name'] = array('Volvo','Dukati','KTR','BMW'); with $_POST['name'] = array($data['data']); ? I ommit, do it you

Michael Over a year ago

@Reza M I want to insert that in one field - when you insert need to be $value=$name1.','.$name2.','.$name3; INSERT INTO table_name (name) VALUES ('$value') for example

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