insert into database

Question

don't give up 0 Junior Poster in Training

12 Years Ago

hello
i want if enter account save in database :)
this is my code , but it's not save anything in database :"(
i don't know why !
i have database "accounts" have "2 row name & account"

<html>
<body>
<?php

$con=mysql_connect("localhost","root","");  

  if(!$con)  
  {  
   die('can not connect'.mysql_error());  
  } 
  mysql_select_db("scms", $con ); 

   ?>

    <form action='' method='POST'>
        <table BORDER="0">
            <tr>
                <td>facbook account :</td>
                <td><input type="text" name="facebook" maxlength="50"/></td>
                <td><input type="SUBMIT" name="save1" value="save"/></td>

              </tr>

              <tr>
                <td>twitter account :</td>
                <td><input type="text" name="twitter" maxlength="50"/></td>
                <td><input type="SUBMIT" name="save2" value="save"/></td>
                 </tr>
                 </table>
                   </form>
                <?php

if(name=="save1")
{
$accout_f = $_POST[facebook];

 mysql_query("INSERT INTO  accounts (account) VALUES ('$accout_f') WHERE name=facebook");

}
elseif(name=="save2")
{
    $accout_t = $_POST[twitter];

   mysql_query("INSERT INTO  accounts (account) VALUES ('$accout_t') WHERE name=twitter");
 }
?>








</body>
</html>

php

3 Contributors
16 Replies
177 Views
1 Day Discussion Span
Latest Post 12 Years Ago Latest Post by don't give up

diafol

12 Years Ago

This should work

if(isset($_POST['name']) && $_POST['name']=="save1")
{
    $accout_f = mysql_real_escape_string($_POST['facebook']);
    mysql_query("INSERT INTO  accounts (account) VALUES ('$accout_f') WHERE name='facebook'");
}
    elseif(isset($_POST['name']) && $_POST['name']=="save2")
{
    $accout_t = mysql_real_escape_string($_POST['twitter']);
    mysql_query("INSERT INTO  accounts (account) VALUES ('$accout_t') WHERE name='twitter'");
 }

but it's not the way that I'd go about it. Your form means that you can only add one field at a time. It may be better if an user could add both on a single button click.

diafol

12 Years Ago

OK, my mistake:

if(isset($_POST['save1']))
{
$accout_f = mysql_real_escape_string($_POST['facebook']);
mysql_query("INSERT INTO  accounts (account) VALUES ('$accout_f') WHERE name='facebook'");
}
elseif(isset($_POST['save2']))
{
$accout_t = mysql_real_escape_string($_POST['twitter']);
mysql_query("INSERT INTO  accounts (account) VALUES ('$accout_t') WHERE name='twitter'");
 }

diafol

12 Years Ago

Have you echoed anything in the code to see where you're going wrong?

e.g. echo $_POST['facebook'];etc?

Also you you just use an 'insert and go'. Add some error handling.

diafol

12 Years Ago

This will happen before you submit the form as the 'sent' vars don't exist yet. Does it still happen after submit?

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don't give up 0 Junior Poster in Training · Answer 1 · 2012-04-28T09:49:18+00:00

thank you , but it's not work

<html>
<body>
<?php

$con=mysql_connect("localhost","root","");  

  if(!$con)  
  {  
   die('can not connect'.mysql_error());  
  } 
  mysql_select_db("scms", $con ); 

   ?>

    <form action='' method='POST'>
        <table BORDER="0">
            <tr>
                <td>facbook account :</td>
                <td><input type="text" name="facebook" maxlength="50"/></td>
                <td><input type="SUBMIT" name="save1" value="save"/></td>

              </tr>

              <tr>
                <td>twitter account :</td>
                <td><input type="text" name="twitter" maxlength="50"/></td>
                <td><input type="SUBMIT" name="save2" value="save"/></td>
                 </tr>
                 </table>
                   </form>
  <?php

if(isset($_POST['name']) && $_POST['name']=="save1")
{   
 $accout_f = mysql_real_escape_string($_POST['facebook']);
 mysql_query("INSERT INTO  accounts (account) VALUES ('$accout_f') WHERE name='facebook'");
 }    
 elseif(isset($_POST['name']) && $_POST['name']=="save2")
 {    $accout_t = mysql_real_escape_string($_POST['twitter']); 
    mysql_query("INSERT INTO  accounts (account) VALUES ('$accout_t') WHERE name='twitter'");
}
?>


</body>
</html>

don't give up 0 Junior Poster in Training · Answer 2 · 2012-04-28T15:28:25+00:00

don't give up 0 Junior Poster in Training

12 Years Ago

not save anything :""(

don't give up 0 Junior Poster in Training · Answer 3 · 2012-04-28T16:24:54+00:00

i try this code

<html>
<body>
<?php

$con=mysql_connect("localhost","root","");  

  if(!$con)  
  {  
   die('can not connect'.mysql_error());  
  } 
  mysql_select_db("scms", $con ); 
   ?>

    <form action='' method='POST'>
        <table BORDER="0">
            <tr>
                <td>facbook account :</td>
                <td><input type="text" name="facebook1" maxlength="50"/></td>
                <td><input type="SUBMIT" name="save1" value="save"/></td>

              </tr>

                <tr>
                <td>twitter account :</td>
                <td><input type="text" name="twitter1" maxlength="50"/></td>
                <td><input type="SUBMIT" name="save2" value="save"/></td>

              </tr>

                 </table>
                </form>   
  <?php
  echo $_POST['facebook1'];
   echo $_POST['twitter1'];
  ?>

</body>
</html>

this output
error :(

don't give up 0 Junior Poster in Training · Answer 4 · 2012-04-28T20:15:15+00:00

if i write this

$a=$_POST['facebook1'];

 $sql="INSERT INTO  $accounts (account) VALUES ('$a') WHERE name='facebook'";

it's not work

but if write this

 $a=$_POST['facebook1'];

 $sql="INSERT INTO  $accounts (name,account) VALUES ('aaa','$a')";

it's work

why? :"(

veedeoo 474 Junior Poster Featured Poster · Answer 5 · 2012-04-28T20:44:41+00:00

Hi,

I think you need to focus more on validating which submit buttons were click. You have two of them. If I click one only, then on POST, you will get the undefined index error on the other.

Why not try validating your form submission based on the submit button clicks. for example

  if((isset($_POST['save1'])) || (isset($_POST['save']))){

   ## check here which one is not empty, and then prepare it for the database inclusion or posting.


  }

Please forgive my vigorous use of parenthesis. I can't just help it and it is becoming more of a second nature to me overtime, it is probably because of my mathematics background and calculator projects I have been working on and in the past. In mathematics, extra care is needed in isolating integers and variables, so that the program will calculate based on the PEMDAS convention. Otherwise, the calculated output is inaccurate, and others will not be able to replicate it.

don't give up 0 Junior Poster in Training · Answer 6 · 2012-04-29T03:07:52+00:00

don't give up 0 Junior Poster in Training

12 Years Ago

problem it's here

WHERE name='facebook'

diafol · Answer 7 · 2012-04-29T07:08:04+00:00

You don't need a Where clause on an insert query. It doesn't make sense to do so in your example

don't give up 0 Junior Poster in Training · Answer 8 · 2012-04-29T12:55:13+00:00

no, i need if enter account in facebook save in account if name == facebook

don't give up 0 Junior Poster in Training · Answer 9 · 2012-04-29T17:00:09+00:00

i don't know what is problem here
$sql="INSERT INTO $accounts (account) VALUES ('$facebook') WHERE name='facebook'";

if i write same this
$sql="INSERT INTO $accounts (name,account) VALUES ('fgfgfg','$facebook') ";
it's work !!!!

diafol · Answer 10 · 2012-04-29T17:41:00+00:00

Like I said WHERE clause doesn't belong in INSERT query. The WHERE clause is usually linked to SELECT, UPDATE and DELETE. As you are inserting, the WHERE clause has no meaning - not in this context anyway.

don't give up 0 Junior Poster in Training · Answer 11 · 2012-04-29T18:31:24+00:00

don't give up 0 Junior Poster in Training

12 Years Ago

ok , i will try with updata

don't give up 0 Junior Poster in Training · Answer 12 · 2012-04-29T19:32:28+00:00

don't give up 0 Junior Poster in Training

12 Years Ago

it's work , use update