PHP - Storing And Displaying Images With Php/mysql

I have a database with all users.. Within that database all users have an id. How would I make it so users can "friend" other users? Would I need to make a new table for that?

Im sorry for the dumb question.. I havent dealt with php or mysql in about two years because I was busy getting engaged and all.. But now that I have time again I think im gonna take up one of my old projects. Granted I probably could have answered this question myself back then but now im finding some trouble taking up programming again. :\

Here is my code...

Code: [Select]
<?php
mysql_connect("localhost", "user", "pass")or die("cannot connect"); 
mysql_select_db("database")or die("cannot select DB");
 
$myemail = mysql_real_escape_string($_POST['myemail']);  
$mypassword = mysql_real_escape_string($_POST['mypassword']); 
$mypassword = md5($mypassword);

$myemail = stripslashes($myemail);
$mypassword = stripslashes($mypassword);

$sql="SELECT * FROM users WHERE email='$myemail' and password='$mypassword'";
$result=mysql_query($sql);

$count=mysql_num_rows($result);

if($count==1){
session_start();  
$_SESSION["myemail"]= "$myemail"; 
header("location:home.php");
}
else {
header("location:fail.php");
}
?>
What can I do to this code so that it will also store first name from the database inside a session?

Hi

I have an array of checkboxes whose values if checked can be updated in mysql. The code I have below accomplishes that just fine.


On my form I have:


print "<form method='post' action='update.php'>\n";

/////// mysQL query here

$myID = $itemRow['myID'];
$chk =  $itemRow['item_shipped'] == 1 ? 'checked' : '';

echo "<input type=checkbox name=cbox[] value='{$itemRow['myID']}' $chk>"; 

echo"</form>"; 




The above code displays various items with a checkbox next to them. So if that checkbox is checked the code below stores that in mysql.


On my update.php page I have:




if(sizeof($_POST['cbox'])) {

//means if at least one check box is selected

foreach($_POST['cbox'] AS $id) { 

mysql_query("UPDATE om SET item_shipped ='1' WHERE myID=$id"); 
} //end foreach

} //end IF 







The problem is though i can check a checkbox and store that value '1' in mysql, I have no way of unchecking an already checked checkbox and storing the '0' in mysql.

Can anyone help?
Thanks in advance

I need some help with this. A user fills out a form, one of the fields is a zip code field. I need to retrieve that value from MySQL store as a session var and set that value as a variable to use with a weather display API. The ID is being stored from the form page.

Here is what I have so far, after the values are submitted into the DB.


<?php
session_start();
$con = mysql_connect("localhost","peter","abc123");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("my_db", $con);

$result = mysql_query("SELECT * FROM Profile WHERE id='{$_SESSION['id']}");

while ($row = mysql_fetch_assoc($result)) {
    $_SESSION['id'] = $row['id'];
$_SESSION['zip'] = $row['zip'];

}


mysql_close($con);
?>



and then for the weather API, I need to set the stored variable to something

$zip = 'stored zip code value'; 

I've got an array like this:

$firstquarter = array('January', 'February', 'March');

Then I'm adding it to a MySQL DB as follows:

$insert = "INSERT INTO sometable (months) VALUES ('$firstquarter') ";
$insertresult = mysql_query($insert) or die ('Error in Insert query: ' . mysql_error());

When I use PHPMyAdmin I see a this value stored in the DB:

'Array'

Can I retrieve the values, by index, from this database field??

Hey, i need help storing an image in my database via the URL(image location) at the moment my php code is storing the image in a folder on the directory called upload. here is the code:
<?php
// Where the file is going to be placed
$target_path = "upload
/";

/* Add the original filename to our target path. 
Result is "uploads/filename.extension" */
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
$target_path = "upload/";

$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);

if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
    echo "The file ".  basename $_FILES['uploadedfile']['name']).
    " has been uploaded";
} else{
    echo "There was an error uploading the file, please try again!";
}
?>

Click <a href="products.php">HERE</a> to go back to form


if someone could help me i'd be very grateful

Hello everyone,   I have a card database and everything works perfectly besides one thing.. I can't store 

'

 in my database via my form, i however do can store them into the database via PHPMyAdmin. I dont know what i've been doing wrong and it really bothers me. If any of you guys could help me out. Here's all the code you would need to find the issue.   This is the form file

<!doctype html>
<html>
<head>
<meta charset="utf-8">
<title>Edit Page</title>
</head>
<body>
<h1 align="center"> Add Cards</h1>
<form action="insert.php" method="POST">
      <input type="text" name="name" placeholder="Name" />
      <input type="text" name="color" placeholder="Color" />
      <input type="text" name="type" placeholder="Type" />
      <input type="text" name="subtype" placeholder="Sub Type" />
      <input type="text" name="power" placeholder="Power" />
      <input type="text" name="toughness" placeholder="Toughness" />
      <br>
      <input type="text" name="manacost" placeholder="Converted Mana Cost" />
      <input type="text" name="rarity" placeholder="Rarity" />
      <input type="text" name="expansion" placeholder="Expansion" />
      <input type="text" name="foil" placeholder="Foil" />
      <input type="text" name="stock" placeholder="Stock" />
      <input type="submit" value="Save" />
	  
</form>
</body>
</html> 

This inserts it into my database.

<?php 

($GLOBALS["___mysqli_ston"] = mysqli_connect("",  "",  "", , ))or die("cannot connect");  
((bool)mysqli_query($GLOBALS["___mysqli_ston"], "USE e_industries"))or die("cannot select DB"); 

$name = $_POST['name']; 
$color = $_POST['color']; 
$type = $_POST['type']; 
$subtype = $_POST['subtype']; 
$power = $_POST['power']; 
$toughness = $_POST['toughness']; 
$manacost = $_POST['manacost']; 
$rarity = $_POST['rarity']; 
$expansion = $_POST['expansion']; 
$foil = $_POST['foil']; 
$stock = $_POST['stock']; 

$sql="INSERT INTO Osiris (Name, Color, Type, Subtype, Power, Toughness, Manacost, Rarity, Expansion, Foil, Stock) 
VALUES ('$name', '$color', '$type', '$subtype', '$power', '$toughness', '$manacost', '$rarity', '$expansion', '$foil', '$stock')"; 
$result=mysqli_query($GLOBALS["___mysqli_ston"], $sql); 

if($result){ 
echo "Successful"; 
echo "<BR>"; 
echo "<a href='add.html'>Back to main page</a>"; 
} 

else { 
echo "ERROR"; 
} 
?>

If anyone could help me out that would be great!
Edited by OsirisElKeleni, 05 October 2014 - 12:29 PM.

I am working on a kind of CMS for my own website which no one else will be using but me as a way of improving my php skills, and am having problems with retrieving data from the database that holds both text and php code. I have searched the web and found that i should be using eval() for the code to be executed before it is send to the browser but cannot get it to work and can't find my mistake(s).

the php code will always be the same, and is supposed to retrieve the id number of a page to use in a link (and works fine when tested by loading the code directly without retrieving it from the database)

this is an example of data stored in the database
Code: [Select]
The <a href="page_builder.php?id=<?php echo $page->id('mines_DwarvenMines') ?>">Dwarven Mines</a> have a great selection of Ores,...
Of course when I leave it like this, hovering over the link in my page will show exactly that and lead to nowhere
Code: [Select]
localhost/page_builder.php?id=<?php echo $page->id('mines_DwarvenMines') ?>
Most of these links appear in tips given at the end of the page and are processed as followed
Code: [Select]
$questtips = $quest->getQuestTips();
$tips = "";
if ($questtips == "none")
{
$tips = "/";
}
else
{
foreach($questtips as $tip)
{
$tips .= "<li>";
$tips .= $tip->getTip();
$tips .= "</li>";
}
}

and finally put on screen by the presentation layer as followed
Code: [Select]
   <h2>Tips &amp; Extra Info</h2>
   <div class="tipsList">
      <ul>
   <?php echo $tips ?>
      </ul>
   </div>
I have tried all sorts to get the code to be executed when retrieved from the database before being send to the browser so that this particular link would say "localhost/page_builder.php?id=57" but I cannot get it to work, though I suspect it is fairly easy.

I suspect I would have to store the data in a different format in the database? And how exactly do I use the eval() function in my case?
Could someone please adjust my code so that it does work?

Thanks

Hello!

I am designing a user registration form which has some radio buttons for people to select. I am confused as to how I should store the value selected by the user into the database.

Here is my code, just in case.
<li id="li_2" >
<label class="description" for="element_2">I am a  </label>
<span>
<input id="element_2_1" name="element_2" class="element radio" type="radio" value="1" />
<label class="choice" for="element_2_1">Student</label>
<input id="element_2_2" name="element_2" class="element radio" type="radio" value="2" />
<label class="choice" for="element_2_2">Staff</label>
<input id="element_2_3" name="element_2" class="element radio" type="radio" value="3" />
<label class="choice" for="element_2_3">Member</label>
<input id="element_2_4" name="element_2" class="element radio" type="radio" value="4" />
<label class="choice" for="element_2_4">Other</label>
</span>
</li>   

helllo dear php-commmunity


new to Ruby - i need some advices -

i plan to do some requests in osm-files. (openstreetmap)

Question - how can i store the results on a Database -
eg mysql or - (if you prefer postgresql) -

note: my favorite db - at least at the moment is mysql


here the code

    require 'open-uri'
    require "net/http"
    require 'rexml/document'
     
    def query_overpass(object_type, left,bottom,right,top, key, value)
       base_url = "http://www.overpass-api.de/api/xapi?"
       query_string = "#{object_type}[bbox=#{left},#{bottom},#{right},#{top}][#{key}=#{value}]"
       url = "#{base_url}#{URI.encode(query_string)}"
       resp = Net::HTTP.get_response(URI.parse(url))
       data = resp.body
       return data
    end
     
    overpass_result = REXML::Document.new(query_overpass("node", 7.1,51.2,7.2,51.3,"amenity","restaurant|pub|ice_cream|food_court|fast_food|cafe|biergarten|bar|bakery|steak|pasta|pizza|sushi|asia|nightclub"))
     
    overpass_result.elements.each('osm/node') {|x|
      if !x.elements["tag[@k='name']"].nil?
        print x.elements["tag[@k='name']"].attributes["v"]
      end
      print " | "
     
      if !x.elements["tag[@k='addr:postcode']"].nil?
        print x.elements["tag[@k='addr:postcode']"].attributes["v"]
        print ", "
      end
      if !x.elements["tag[@k='addr:city']"].nil?
        print x.elements["tag[@k='addr:city']"].attributes["v"]
        print ", "
      end
      if !x.elements["tag[@k='addr:street']"].nil?
        print x.elements["tag[@k='addr:street']"].attributes["v"]
        print ", "
      end
      if !x.elements["tag[@k='addr:housenumber']"].nil?
        print x.elements["tag[@k='addr:housenumber']"].attributes["v"]
      end
      print " | "
      print x.attributes["lat"]
      print " | "
      print x.attributes["lon"]
      print " | "
      if !x.elements["tag[@k='website']"].nil?
        print x.elements["tag[@k='website']"].attributes["v"]
      end
      print " | "
      if !x.elements["tag[@k='amenity']"].nil?
        print x.elements["tag[@k='amenity']"].attributes["v"]
        print " | "
      end
      puts
    }

look forward to hear from you

again - i would love to store it on a mysql - database - if possible. If you would prefer postgresql - then i would
takte this one.... ;-)



look forward to hear from you

again - i would love to store it on a mysql - database - if possible. If you would prefer postgresql - then i would
takte this one.... ;-)

well - i guess that the answer to this will be the same no matter what language we are using.
If the db is a sql database we need to design the database schema and create the tables in the database.

The first step in accessing a db in our code is to get a connection to it.
If ruby is our choice of language, a search for "ruby sql connector" will give us
lots of options to read about.

Well - we also can do it in PHP. What do you think!?

Next, based on the schema we have designed, we need to create queries suitable for storing the data.
We will likely need to consider our transactional model.

Again, searching "ruby sql transactional model" will give us plenty of food for thought. Finally, we may want or need to close the connection to the database.

i have a dictionary  - THIS IS obviously a python dictionary

an this has approx 8 000 lines with records


[/CODE]

$ python printer.py
{'url': 'http://www.site1_com' 'cname': 'butcher', 'name': 'cheng', 'email': 'mail1@hotmail.com'}
{'url': 'http://www.site2_com' 'cname': 'dilbert', 'name': 'James', 'email': 'mail2@hotmail.com'}

[code=auto:0]i have a mysql-db up and runing in my opensuse

there i have created a db with the fields

url
cname
name
email



i use the

import MySQLdb

i studied this documentation he http://stackoverflow...abase-in-python

but i think this goes a bit over my head.


- well how can i get the data  ( in other words the dictionary) into the database?

love to hear from you

greetings

I created a simple search box which will query my table and match the input value to one of my columns. two of these columns store comma separated values. if i query a column other than a column which stores my csv i can see my search results. if i query a column which stores my csv i will not see results unless the search value matches the first value within the column. how would i be able to get say the second or third or forth value. here is the code i am using to query the table any help would be appreciated thanks.

Code: [Select]
$q = $this->db->query("SELECT * FROM table WHERE col1 LIKE 'searchvalue'".
             " OR col2 LIKE 'searchvalue'".
             " OR col3 LIKE 'searchvalue'".
             " OR col4 LIKE 'searchvalue'".
             " OR FIND_IN_SET('searchvalue', col5) > 0 ".
             " OR FIND_IN_SET('searchvalue', col6) > 0");

So I am trying to display an image from a database (I know it's not the best idea but it's what I was told to do). Anyway I'm having problems displaying it. Here is the code for displaying the image.
$cat = $_GET['cat'];

include 'includes/openDbConn.php';

$sql = 'SELECT * FROM CushionsCategories WHERE CushionCategory = "'.$cat.'"';

echo $sql;

$result = mysql_query($sql); 

echo '<table>';

while($val = mysql_fetch_assoc($result)){

$cush = 'SELECT * FROM Cushion WHERE SKU = "'.$val['CushionSKU'].'"';

echo $cush;

$cushres = mysql_query($cush);

$cushval = mysql_fetch_assoc;

$img = 'SELECT * FROM Images WHERE SKU = "'.$val['CushionSKU'].'"';

echo $img;

$imgres = mysql_query($img);

$imgval = mysql_fetch_assoc($imgres);

if( $i % 3 == 0 ) {

      echo '</tr><tr>';
  

}

echo '<td><img src="image.php?sku='.$val['CushionSKU'].'" name="'.$imgval['FileName'].'" description="'.$imgval['Description'].'" /></td>';

}

echo '</tr></table>';

and the page that gets the image:
$sql = 'SELECT Image FROM Images WHERE SKU = "'.$sku.'"';
//echo $sql;
 $result = mysql_query($sql) or die("Invalid query: " . mysql_error());
 
        // set the header for the image
        header("Content-type: image/jpeg");
        echo mysql_result($result, 0);

Thanks for any help in advance.

Hello I am using this script right here
http://www.nearby.org.uk/sphinx/search-example5-withcomments.phps
and am trying to show images from my db but I can only get it to show the first image. Does anyone know how to make it show all images if images even exist for that particular search result?

I changed the query to $CONF['mysql_query'] = '
SELECT l.link_id AS id, l.title AS title, l.fulltxt AS body, l.url AS url, m.media_id AS im_id, m.title AS im_title, m.thumb_link AS im_t_link
FROM search1_links AS l LEFT JOIN search1_media AS m ON (m.link_id = l.link_id)
WHERE l.link_id IN ($ids)
';

and added

if ($row['im_id']) {
echo '<img src="'.($row['im_t_link']).'" height="100px" width="100px"> ';
}
right here

            //Actully display the Results
            print "<ol class=\"results\" start=\"".($currentOffset+1)."\">";
            foreach ($ids as $c => $id) {
                $row = $rows[$id];
                
                $link = htmlentities(str_replace('$id',$row['id'],$CONF['link_format']));
                print "<li><a href=\"$link\">".htmlentities($row['title'])."</a><br/>";
                
                if ($CONF['body'] == 'excerpt' && !empty($reply[$c]))
                    print ($reply[$c])."</li>";
                else

           if ($row['im_id']) {
echo '<img src="'.($row['im_t_link']).'" height="100px" width="100px"> ';
}

                    print htmlentities($row['body'])."</li>";
            }
            print "</ol>";
            
            if ($numberOfPages > 1) {
                print "<p class='pages'>Page $currentPage of $numberOfPages. ";
                printf("Result %d..%d of %d. ",($currentOffset)+1,min(($currentOffset)+$CONF['page_size'],$resultCount),$resultCount);
                print pagesString($currentPage,$numberOfPages)."</p>";
            }

Can anyone help me to be able to display all images if they exist? Thanks.