PHP IF statement not taking variable into account! - php

I have a tabled view in a while loop, where a user can view information on books.
For example, book ISBN, book name, read status...
Basically, when the user sets their 'readstatus' to 'complete' I want that specific table row to become grey! The logic is very straight forward, however I can't get my IF statement to recognise this:
if ($readstatus == 'complete') {
echo '<tr class="completed">';
}
else if ($readstatus != 'complete') {
echo '<tr class="reading">';
}
I'm obviously doing something wrong here, table content to change if the value of 'readstatus' = 'complete', if not, then output is the default

Why are you using $_GET? Does this information come from an HTML form or a URL etc... ?
I suspect you meant to change $readstatus = $_GET['readstatus']; to $readstatus = $row['readstatus'];.
$_GET is an aray of GET parameters which come from the query string.
$row is a row in your database, so if the information is in the database - which I suspect it is - you want to use $row instead of $_GET.

Try changing $readstatus = $_GET['readstatus']; to $readstatus = $row['readstatus'];

The $_GET function relies on the value being contained in the query string of the URL, and it has nothing to do with the database. I have a hunch you're trying to get the value from the database here and you're using the wrong function to do it.

$_GET['readstatus'] says the value is coming from the browser.
$row['readstatus'] says the value is coming from the database.
You need to decide which should take precedence-- probably the $_GET['readstatus']` because it's what the user wants to change. If that's the case, you need to update your database with the new readstatus before you requery the db for the dataset.

Related

How to store post variables value

I got a Index page on which search page is included, and when I submit it, it passes values to find.php through action and method post. The code is below
if($_POST['searchsubmit']=="Search"){
$cat=$_POST['searchcategory'];
$area=$_POST['searcharea'];
$term=$_POST['searchbox'];
}
The above code is written on find.php, Now when I try to implement paging through basic paging method with where conditions to make appropiate search query
$where="where approved='yes'";
if($term!=""){
$where.=" and name like '%$term%'";
}
if($cat!=""){
$where.=" and category like '%$cat%'";
}
if($area!=""){
$where.=" and area like '%$area%'";
}
$start=0;
$end=5;
if($_GET['page']!="")
{
$start=$_GET['page']*$end;
}
Where $start is my initial limit, and $end is my number of records. For the first page of paging, I pass a variable page with 0 for first page
First
and my search query now becomes
$que="select * from shops ".$where." ORDER BY likes DESC limit $start,$end";
As soon as I click on "first", My new link become "/find.php?page=0"
and the post values which I recivied from index page search bar are lost.
Is there any way to retain those values ?The two methods which I though are sending them again through url with GET, or the other way is to store them in session.
Is there any third method available ?
Marc is absolutely right. Do not use the code as it is.
As an alternate solution to your problem -
Your page index.php (search form) submits to itself
Assemble your search query as querystring in index.php if its a post
Redirect to find.php with the assembled querystring
Every search information will always be in the querystring.
Use your pagination happily.
The comments are correct.
Use:
// Start the session
session_start();
// Save variables into session
$_SESSION['somevalue'] = $_POST['value'];
Then when any page calls session_start it will have access to $_SESSION['somevalue']
Also, you are wide open for SQL injection. Sanitize your values to ensure no one can put arbitrary sql code into the string. if you are using mysqli it should as simple as this:
// After connecting to the DB
$_POST['somevalue' = $mysqli->real_escape_string($_POST['somevalue']);
Then be sure to hardcode quotes around string values like you are doing.
If you want to be safer you can use prepared statement instead.
Hope this helps.

Search with PHP/MySQL and dynamically update with jQuery

I was wondering how to do the following best with PHP/MySQL and jQuery:
There is a basic search mask where you enter a city and a from-to-date. You process to the search-result page, where you then can narrow your search results with certain parameters (checkboxes, jQuery slider, text-input, ...). The search-results should then update on the fly without the whole page being reloaded...
I manage to use jQuery ajax and load to send information to another php file, perform e.g. a SELECT and return the results to the search detail page, but I don't know how to combine different changes that narrow the search results.
Furthermore, there are already results on the detail page, so I do not need to add more results but "delete" the results that do not fit anymore...
The thing is that each parameter to narrow the search is connected to another table in the database. Do I have to and how do I add joins to the original query...? Or am I thinking in the wrong direction?
Yes, this is absolutely the right direction. Use
$(document).ready(function() {
$('#ID_OF_YOUR_ELEMENT_TO_LOAD_INTO').load("load.php?parameter1=<?php echo $parameter1; ?>&parameter2=<?php echo $parameter2; ?>");
});
to get the results when the user gets on the page for the first time, to get the results according to your city and your dates.
Check in the load.php which parameters are set and use the ones that are set to build your query. Then, when the form (or forms, depending) are updated, you have to use .load again, like this:
$('#ID_OF_YOUR_FORM_BEING_UPDATED').change(function() {
$('#ID_OF_YOUR_ELEMENT_TO_LOAD_INTO').load("load.php?parameter1=<?php echo $parameter1; ?>&parameter2=<?php echo $parameter2; ?>&parameter3=<?php echo $parameter3; ?>");
});
Get the initial tuples via PHP/MySQL, save them into some Javascript structure and create the html needed to display the data with javascript from this structure.
Any time you want to filter the data you rewrite the html and check the filter condition on the fly, e.g. don't write tuples from the structure that don't match your filter condition.
You can see who this is done at http://www.wowhead.com
This is of course just one way. ;-)
You could always write some code to generate an SQL query based on passed arguments.
You ajax could query the page with a bunch of arguments in addition to your basic city and from-to date based on what the user has selected. If your page preserves the previous search options selected, it should be able to just let the user add on more options and keep processing them in the same way. Your php would then test to see if the arguments are set in the $_POST or $_GET variable ($_POST is more secure for ajax generally, but my example will use $_GET for simplicity) and build the query like that.
Example:
Javascript generates a query like searchAjaxHandler.php?city=Chicago&from=2012-03-01&to=2012-03-05&someColumnLowerRange=500&someColumnUpperRange=700
Your php script then processes as follows:
$query = "SELECT * FROM Data WHERE City=? AND Date > ? AND Date < ?";
$arguments = array($_GET['city'], $_GET['from'], $_GET['to']);
if (isset($_GET['someColumnLowerRange'])) {
$query .= " AND someColumn > ?";
$arguments[] = $_GET['someColumnLowerRange'];
}
if (isset($_GET['someColumnUpperRange'])) {
$query .= " AND someColumn < ?";
$arguments[] = $_GET['someColumnUpperRange'];
}
//execute the query
//using PDOs (google them...they are a good way to prevent sql injection and
//support multiple database types without modifying code too much), create a
//statement with the above query in put the statement in $statement
$statement->execute($arguments); //this uses the $arguments array to fill in the prepared statement's ?'s
//then do the stuff to get the retrieved rows out of the result returned
After all that, the javascript side would just to the same thing you were doing before by replacing all the previous results with the results that you got back.

Returning a resultset from one page to another in PHP

I have 2 php pages: query.php and result.php.
In query.php, I am executing a query (select) statement. It's returning a resultset
$rs = mysql_query($query);
Now I want to return this resultset from query.php to another page result.php and work with it. Like this:
In query.php:
return $rs
and in result.php:
$result = executeQuery($query) // we get the resultset in this variable
while ($row == mysql_fetch_array($result){
//do something
}
If the above is not recommended, please provide me with alternatives. But I want the query function and resultset in different pages.
You could just include results.php in your query.php page if you're just looking to keep the code separate in the source files but aren't actually required to redirect from one page to another:
In query.php:
$rs = mysql_query($query);
include "results.php";
In results.php:
while ($row == mysql_fetch_array($rs){
//do something
}
As far as trying to "return $rs" from one page to another that's not how PHP works. The return statement is only valid within a function. If you want to actually pass data from one PHP page to another and will be redirecting to that other page then you'll need to use either a session, a cookie, pass it in the URL (i.e. use GET) or use curl and add it as a POST var.
If this is really the way it must be, store the result set in a database somewhere or in a file and give each result a unique name. Then pass that name to the next page so it can be retrieved.
query.php will redirect to result.php?result_set=ab24sdfsdfklls for instance.
This has the added advantage that you can use the result_set as often as you want. Visitors can have multiple result sets during one visit. They can share the URL of the result set page with other people, etc.
Just be sure to eventually prune the data store as it will just keep on growing, but that's another matter entirely.

add a name value pair to posted data for joomla to consume and insert into DB

I'm trying to use the Joomla framework to make a create in the main content table.[http://docs.joomla.org/How_to_use_the_JTable_class] This works fine except that some data comes from posted variables and some from logic that happens when a file is uploaded moments before (store the random image name of a jpg)
$data=&JRequest::get('post');
this takes a ref to the posted values and I want to add to this Array or Object my field. The code I have makes the new record but the column images, doesnt get my string inserted.
I am trying to do something like$data=&JRequest::get('post');
$newdata=(array)$data;
array_push($newdata,"images"=>"Dog");
i make newdata as data is a ref to the posted variables and i suspect wont there fore allow me to add values to $data.
I'm a flash guy normally not a php and my knowledge is letting me down here.
Thanks for any help
Right, first thing:
$data=&JRequest::get('post');
$data is an array, you do not have to cast it. To add another element to the array as described in the comments do this:
$data['images'] = 'cats';
If you are using normal SQL to do the insert then you would do something like this to get the last inserted id e.g. the id of the row you just inserted:
$db = $this->getDBO();
$query = 'Some sql';
$db->setQuery($query);
if (!$db->query()) {
JError::raiseWarning(100, 'Insert failed - '.$db->getErrorMsg());
}
$id = $db->insertid();
If you are developing in Joomla I suggest you use the db functions provided to you rather than mysql_insert_id();
[EDIT]
If you want to use store then you can get the last inserted id like so:
$row->bind($data);
$row->check();
$row->store();
$lastId = $row->id;

redirect/echo if file.php?id=* not found/doesn't exist

Hey guys,
PHP and MySQL newguy here. Wrote this php file which display the content of a row relative to the ID stated in the URL ( eg row 3 is file.php?id=3 ), heres the source: http://pastie.org/1437017
If I goto an id to which the relative row does not exist (eg .php?id=99999999999999), what do I put to in to get it to redirect to another page or echo 'FAIL'. I though about using the if command, but couldn't figure out the syntax. I also looked around the web, but no avail.
Thanks guys
You have the following line:
$name=mysql_result($result,$id,"name");
If there is no row with the id $id, $name will be false. You could therefore do the following:
if (!$name) {
header('Location: http://yoururl.com');
die();
}
Better yet would be to modify your query to this:
$query="SELECT * FROM likes where id=$id";
and then do
if (!$num) {
header('Location: http://yoururl.com');
die();
}
where $num is the number of row returned, as set in your existing code.
Edit As noted elsewhere in this question, it is probably better to serve a 404 Not Found page with appropriate content, rather than redirecting to another page. I can just about imagine a situation where redirection is appropriate, but if your redirection page says "item not found", this is the wrong approach.
I'd redesign your query to something like
SELECT * FROM table WHERE id = $id;
where $id is the $_GET value - sanitised of course.
if that query returns any results (mysql_num_rows($result)==1)
then you know a valid record has been found. If not, the id doesn't exist, so you can throw an error/redirect.
mysql_num_rows() gives you the number of rows in your select, so if that value is 0, you know there isnt any row with that given id.
if (mysql_num_rows($result)==0){
echo "There are no rows with this id";
}else{
// Your normal code
}

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