How would i got about this...

[Lee-Bartlett]

Ok i have my database with 1 table in it, say i wanna edit a record which has been submited, say because it is stupid or typoed etc. I would like to be able to update it with text feilds, say name is spelt wrong, i have a text feild where i typed the correct name in and it updates that row of information. How would i go about doing this, any nice tuts or some advice is great. It can be very simple nothing advanced

[wildteen88]

Your first the pull whatever row you want to edit from the database, you'll place each column within a form field. When you submit the form you just run an UPDATE query, eg

 

UPDATE TABLE `table_here` SET col1name=`$col1value`,  col12name=`$col2value` etc WHERE row_id_col=$rowid

[Lee-Bartlett]

This i what i had so far,

<?php

mysql_select_db("roberts_work", $connect);

mysql_query("UPDATE tblbasicform SET location = 'Harlow'
WHERE name = 'chris' AND email = '[email protected]'");

mysql_close($connect);



?>

 

But i need it to work in a form, for example, i have a db with email and name in, i search the users by email. On my page i would like 2 text feilds and an update button, when i fill out them 2 feileds, what ever user with that name, is updated with the text i put in.

[hcdarkmage]

If you have setup a unique primary key in your database, you could use it for updating the information.

 

Create the form that has the fields you want to update the database:

<form name="form1" method="post">
Input First Name: <input type="text" name="firstname" /><br />
Input Email: <input type="text" name="email" /><br />
Input Location: <input type="text" name="location" /><br />
Input ID Number: <input type="text" name="id" /><br />
<input type="submit" name="submit" value="Update Data" />
</form>

 

Of course being a little rusty, in the PHP section of the page you do something like this:

<?php
  
  foreach($_POST as $key => $value)
      $$key = $value;  //This sets the form field names to values.

  mysql_query("UPDATE table SET location = '$location', name = '$firstname', email = '$email' WHERE id = '$id'") or die(mysql_error());

?>

 

That way you have a clean update of any information that needs to be updated.

 

 

[Lee-Bartlett]

I dont understand that completly, it doesnt update when i changed everything to. I would of thought first i would need a box where i search for name/id number, then if i get a result back it would let me use 4 text boxxes to update the rows i have.

[hcdarkmage]

I dont understand that completly, it doesnt update when i changed everything to. I would of thought first i would need a box where i search for name/id number, then if i get a result back it would let me use 4 text boxxes to update the rows i have.

 

You could do a drop down box that selects all the people in the database and sets the values of the form to auto-populate with the information and you can change and update the information.

 

<?php

  $nameQry = "SELECT * FROM database";
      $result = mysql_query($nameQry) or die(mysql_error());
  
  while($row = mysql_fetch_array($result)){
    $name = $row['name'];
    $email = $row['email'];
    $location = $row['location'];
    $id = $row['id'];
  }

?>

 

Then in your form control you have this:

 

<form name="form1" method="post">
<select id="names" name="names">
  <option>option values</option>
</select>
  Input First Name: <input type="text" name="firstname" value="<? echo $name ?>" /><br />
  Input Email: <input type="text" name="email" value="<? echo $email ?>" /><br />
  Input Location: <input type="text" name="location" value="<? echo $location?>" /><br />
  Input ID Number: <input type="text" name="id" value="<? echo $id?>" /><br />
<input type="submit" name="submit" value="Update Data" />
</form>

 

or something along those lines.

[Lee-Bartlett]

only problem with this is, the drop down stays on option value and the feilds fill straight away with the first row in my db, when i try change it or update it.

[hcdarkmage]

Alright....lets try this:

 

<?php

  $nameQry = "SELECT * FROM database";
      $result = mysql_query($nameQry) or die(mysql_error());

  $i = 0;  

  while($row = mysql_fetch_array($result)){
    $name = $row['name'];
    $email = $row['email'];
    $location = $row['location'];
    $id = $row['id'];
    $optBox[$i] = $row;
    $i++;
  }

  if(isset($names)){
     $valueQry = "SELECT * FROM database WHERE id = '$id'";
         $result = mysql_query($valueQry) or die(mysql_error());

    while($row = mysql_fetch_array($result)){
        $valID = $row['id'];
        $valName = $row['name'];
        $valEmail = $row['email'];
        $valLocation = $row['location'];
    }
  }

?>

 

Then in the form area:

 

<form name="form1" method="post">
<select id="names" name="names">
  <?
     for ($j=0; sizeof( $optBox ); j++){
        echo "<option value='".$optbox[$j]['id']."'>".$optbox[$j]['name']."</option>";
     }
  ?>
</select>
  Input First Name: <input type="text" name="firstname" value="<? echo $valName ?>" /><br />
  Input Email: <input type="text" name="email" value="<? echo $valEmail ?>" /><br />
  Input Location: <input type="text" name="location" value="<? echo $valLocation?>" /><br />
  Input ID Number: <input type="text" name="id" value="<? echo $valID?>" /><br />
<input type="submit" name="submit" value="Update Data" />
</form>

 

This code is untested, but in theory should help.

[Lee-Bartlett]

So close, the drop down has found the database and the form is filled in with the bottom entry of the database, dunno why. All i need now is the update bit.

[Lee-Bartlett]

i forgot to put the form code bit in the code, now im reciving back

 

Parse error: syntax error, unexpected T_INC, expecting ')' in C:\xampp\htdocs\Website\update.php on line 35

 

i took out a ; to see if that fixxed it and i get unexpected 2 string to

[amagondes]

Hi, you are missing the $ infront of your variable j when incrementing it

 

change:

for ($j=0; sizeof( $optBox ); j++){

 

to:

 

for ($j=0; sizeof( $optBox ); $j++){

[hcdarkmage]

Hi, you are missing the $ infront of your variable j when incrementing it

 

change:

for ($j=0; sizeof( $optBox ); j++){

 

to:

 

for ($j=0; sizeof( $optBox ); $j++){

 

Thanks for catching that.  Usually when I code, I go a bit fast and forget to put in the $ and ; every once in a while.  It bugs the crap out of people when that happens, but it happens.