tags:

views:

72

answers:

5

I have a website that has an update query. Example.

Table name -> myTable
Table Content -> id(AI)
                 name  --> myName
                 age   --> 13
                 image --> (no values)
                 birthdate  --> (no values)

I only put values in name and age. In image and birthdate, I haven't put anything.
If I edit it and update it and I don't put anything in name and in age, I only put values in image and in birthdate. The output is like this. (This ouput I wanted)

Table name -> myTable
Table Content -> id(AI)
                 name  --> myName
                 age   --> 13
                 image --> myImage.jpg
                 birthdate  --> Feb. 31, 2010


This is my code:

<?php
  //mysql connect...
  //mysql select database...
$sql = "UPDATE myTable SET name = '".$name."', age = '".$age."', image = '".$image."', birthdate"'$birthdate.."' WHERE id = $id";
mysql_query($sql);
?>

I used this code but the output is:

Table name -> myTable
Table Content -> id(AI)
                 name  --> (no values)
                 age   --> (no values)
                 image --> myImage.jpg
                 birthdate  --> Feb. 31, 2010


Feel free to ask if you don't understand my question
Thank you

A: 

Just have two separate queries:

// update name and age
$sql = "UPDATE myTable SET name = '$name', age = '$age' WHERE id = '$id'";

// Update image and birthday
$sql = "UPDATE myTable SET image = '$image', birthdate = '$birthdate' WHERE id = '$id'";

Also, you can just put all of the variables inside the string since it's double quotes.

Brendan Long
A: 

If you don't want to change a field, don't include it in the update query.

Don Kirkby
+2  A: 

you have an error in your code--

$sql = "UPDATE myTable SET name = '".$name."', age = '".$age."', image = '".$image."', birthdate"'$birthdate.."' WHERE id = $id"

should read:

$sql = "UPDATE myTable SET name = '".$name."', age = '".$age."', image = '".$image."', birthdate = '".$birthdate."' WHERE id = $id"

but that's probably just a typo. Are you sure that the variables are getting set correctly? Try echoing out all the vars to make sure. Also, check that the columns have no default data definition.

julio
Beat me to it: birthdate"'$birthdate.."'
RobertPitt
Thank you. ^^. I forgot the equal sign.
Jordan Pagaduan
+3  A: 

UPDATE leaves columns as they were if you don't include new values in your UPDATE statement.

So you may need to build up the $sql variable with some conditional code that examines what the user entered, and then only include a column if the user gave a value for it.

<?php

$columns = array("name", "age", "image", "birthdate");
$set = array();
foreach ($columns as $col) {
    if (isset($_GET[$col])) {
      $set[] = "$col = '" . mysql_real_escape_string($_GET[$col]) . "'";
    }
}
$sql = "UPDATE myTable "
if ($set) {
    $sql .= " SET " . implode(",", $set);
}
$sql .= " WHERE id = " . (int) $id;
Bill Karwin
Thank you. This is what I need.
Jordan Pagaduan
It might be better to use `!empty($_GET[$col])` instead of `isset`. If the form has fixed inputs (e.g.: inputs are not added or disabled dynamically), then empty inputs would be treated as if a new value was provided.
Justin Johnson
@Justin Johnson: Good point. Season to taste. :-)
Bill Karwin
A: 

Personally i think that quires like this are not that reliable, Especially when it comes down to typecasting etc.

I recommend you try and implement PDO or something.

http://php.net/manual/en/book.pdo.php

Othewise, the error is birthdate'".$birthdate."' witch needs to be birthdate = '".$birthdate."'

RobertPitt