PHP - Assign A Default Value In A Field With A Sql Select Query With Php

How should it look if I want to make a query where a field value should be lower than another field?

Code: [Select]
....WHERE total_points<max_points");

Hi,
I need to call to this text to add a all with value "blank/empty" or "0" to the dropdownlist code below..
it should be inserted int the first position in the dropdownlist and have a value blank and call to this text from the language files:            
            $searchAll = gettext('_LANGUAGE_SEARCH_ALL',_LANGUAGE_SEARCH_ALL,false,false);
            example is:
            ratestitle=>0 gettext('_LANGUAGE_SEARCH_ALL',_LANGUAGE_SEARCH_ALL,false,false);
            the words it calls is "ALL" when the user selects or leaves the "ALL" in the list the value
             remains empty this allows the search to search everything..
basically im stuck how to get it in the dropdown list as the first default position:
can anyone help?

here is my code://this is a search function which calls to the field rate_title the text from the select box is matched to the database table and the search pages shows the search option is the search is successful.. I have 5 of these types of searches but cant give a broader search option to my users as the values must be selected to search--- i need the option "ALL" to have a "EMPTY" value or no value....
         

if (in_array("ratestitle",$searchOptions)&& $showSearchOptions )
            {
            if (empty($sch->filter['ratestitle']) )
               $selectOption=$output['LANGUAGE_SEARCH_RATESTITLE'];
            else
               $selectOption=$sch->filter['ratestitle'];
            $showButton=true;            
         $query = "SELECT DISTINCT rate_title FROM #__rates_table ORDER BY rate_title ASC";
         $dropDownList ="<select class=\"inputbox\" name=\"ratestitle\">";
         $ratesitles =doSelectSql($query);         
         foreach ($ratesitles as $ratetitle)
            {
            $selected="";
            $rate_title=$ratetitle->rate_title;
            if ($ratetitle==".$sch->filter['ratestitle'].")
               $selected="selected";
            $dropDownList .= "<option ".$selected." value=\"".$rate_title."\">".$rate_title."</option>";
            }
         $dropDownList.="</select>";
         $output['RATESTITLE']=$dropDownList;
            }
               

Hello.

I am trying to display only one instance of records that have the same memberid in my db. I am using the following statement but it continues to show all of the records that have the same memberid. Any ideas what I may be doing wrong?

Code: [Select]
$sql = "select DISTINCT memberid, event, category, date, enddate, locality, location, address, city, state, zip, contact, phone, notes, doc1, doc2, doc3, doc4, doc5 from event where date >= '$datenow' ORDER by date ASC";
Thanks for any help!

Hi.  I have two select fields one called category and the other sub category.  Both are in the format.
<select name="category" value="

<?php $qGetCat = "SELECT * FROM club_category" ;

$rGetCat = mysql_query($qGetCat);

while ($Cat = mysql_fetch_assoc($rGetCat))

{

?>" />
                                <option value="<?php echo $Cat['categorys'];?>"><?php echo $Cat['categorys']; ?></option>
                                <?php } ?>
                                </select>
<select name="sub_category" value="

<?php $qGetSubCat = "SELECT * FROM sub_categorys" ;

$rGetSubCat = mysql_query($qGetSubCat);

while ($SubCat = mysql_fetch_assoc($rGetSubCat))

{

?>" />
                                <option value="<?php echo $SubCat['sub_categorys'];?>"><?php echo $SubCat['sub_categorys']; ?></option>
                                <?php } ?>
                                </select>


Now I want to make it so that when a particular category is chosen for example self defence.  The sub category when clicked shows all the self defence stuff.  If watersports was chosen sub category would show things like swimming.  What do I need to look up to do this? At the minute there is no link the way I have coded these to fields.

In the database its self there is a;

category table with categoryID and category_name

sub category table with sub_cat_name and categoryID

I think this will work but I have never done this so its making my head hurt just thinking about it.  Does anyone have any advice, tuts or scripts to do this? 

Thank you

I want to select the user from super_administrators, administrators, teachers and students, and give the user permission based from what table he "came". But is giving the "Query failed" error...

Code: [Select]
<?php

//Start session

session_start();



//Include database connection details

require_once('../config/config.php');



//Array to store validation errors

$errmsg_arr = array();



//Validation error flag

$errflag = false;



//Connect to mysql server

$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

if(!$link) {

die('Failed to connect to server: ' . mysql_error());

}



//Select database

$db = mysql_select_db(DB_DATABASE);

if(!$db) {

die("Unable to select database");

}



//Function to sanitize values received from the form. Prevents SQL injection

function clean($str) {



$str = @trim($str);



if(get_magic_quotes_gpc()) {

$str = stripslashes($str);

}



return mysql_real_escape_string($str);

}



//Sanitize the POST values

$email = clean($_POST['email']);

$password = clean($_POST['password']);



//Input Validations

if($email == '') {

$errmsg_arr[] = 'O campo Email nao foi preenchido.';

$errflag = true;

}



if($password == '') {

$errmsg_arr[] = 'O campo Senha nao foi preenchido.';

$errflag = true;

}



//If there are input validations, redirect back to the login form

if($errflag) {

$_SESSION['ERRMSG_ARR'] = $errmsg_arr;

session_write_close();

header("location: ../index.php");

exit();

}



//Create query

$qry = "SELECT * FROM super_administrators,administrators,teachers,students WHERE email = '$email' AND passwd = '".md5($_POST['password'])."'";

$result = mysql_query($qry);

$member = mysql_fetch_assoc($result);

$table = mysql_field_table('$result', '0');



//Check whether the query was successful or not



if($result) {



if(mysql_num_rows($result) == 1) {



if($table == 'super_administrators')  {

session_regenerate_id();

$_SESSION['SESS_ID'] = $member['id'];

$_SESSION['SESS_NAME'] = $member['name'];

$_SESSION['SESS_EMAIL'] = $member['email'];

session_write_close();

header("location: ../users/sadmin/index.php");

exit();

}



if($table == 'administrators')  {

session_regenerate_id();

$_SESSION['SESS_ID'] = $member['id'];

$_SESSION['SESS_SCHOOL_ID'] = $member['school_id'];

$_SESSION['SESS_NAME'] = $member['name'];

$_SESSION['SESS_EMAIL'] = $member['email'];

session_write_close();

header("location: ../users/admin/index.php");

exit();

}



if($table == 'teachers')  {

session_regenerate_id();

$_SESSION['SESS_ID'] = $member['id'];

$_SESSION['SESS_SCHOOL_ID'] = $member['school_id'];

$_SESSION['SESS_NAME'] = $member['name'];

$_SESSION['SESS_EMAIL'] = $member['email'];

session_write_close();

header("location: ../users/prof/index.php");

exit();

}



if($table == 'students')  {

session_regenerate_id();

$_SESSION['SESS_ID'] = $member['id'];

$_SESSION['SESS_SCHOOL_ID'] = $member['school_id'];

$_SESSION['SESS_CLASS_ID'] = $member['class_id'];

$_SESSION['SESS_NAME'] = $member['name'];

$_SESSION['SESS_REGISTRATION'] = $member['registration'];

$_SESSION['SESS_EMAIL'] = $member['email'];

session_write_close();

header("location: ../users/aluno/index.php");

exit();

}



} else {



$errmsg_arr[] = 'Suas informacoes de login estao incorreta. Por favor, tente novamente.';

$errflag = true;

$_SESSION['ERRMSG_ARR'] = $errmsg_arr;

session_write_close();

header("location: ../index.php");

exit();

}



} else {

die("Query failed");

}

?>

Can anybody clarify the usefulness of "AS" and elaborate on how it operates.

If I select a 'bananas' AS 'yeelow fruit' will it change the TITLE in the table?  Or the table that is being viewed?
If I dump a list of 'FRUITS' into and HTML table, will using AS rename each column for me, or is that handled by MY coing of the HTML table?

Hi Guys,

Can someone please help me before I kill myself!

I have a table holding details of images to be displayed in a gallery.
The table has columns as 'id', 'image_caption', 'file_path', and 'userid'.

The 'id' column is the unique key, and the 'userid' column is obviously id of the user who uploaded the image.

What I would like to do is select and display one image from each unique userid and then display each image along with the details in my gallery. It doesn't really matter which users image is selected aslong as there is only one for each unique user.

I have been trying a number of ways to do this (DISTINCT / subqueries) but I just cant get anything to work.
Could someone please advise me on how I could get this to work.

Cheers in advance.

Hello, i am trying to get my form to list usernames ascending as an option, but the select field just returns blank.

Here is my code:

Code: [Select]
/**
 * editUserForm - Displays the users database table in
 * a nicely formatted field table.
 */

function editUserForm(){
   $q = "SELECT * "
       ."FROM ".TBL_USERS." ORDER BY username ASC";
   $result = $database->query($q);

while ($row = mysql_fetch_assoc($result))
{
echo '<option value="'.$row["username"].'">'.$row["username"].'</option>';
}
}

Code: [Select]
              <form action="/" method="post">

                <div class="column">
                  <p>
                    <select name="user" id="user" placeholder="Select A User To Edit" class="{validate:{required:true}}">
                      <option>Select A User To Edit</option>
                      <? editUserForm(); ?>
                    </select>
                  </p>

                <div class="action_bar">
                  <input type="submit" class="button blue" value="Submit Post" />
                  <a href="#modal" class="modal button">Cancel</a>
                </div>

              </form>

All help is appreciated

Hi guys,
sorry...i'm a newbiiii and actually not that firm with php.
i create some formualrs using while loop.
the forms and input fields receives their names including a variable out of the database.

by submitting to another page almost all values are transmitted except the value of the select field.

I'm acutally using that ( it's just a shortcut ):

 

<form id='contact-form".$KitchConfigID."' name='contact-form".$KitchConfigID."' action='_pages/_kitchenconfig/adm_saveKitchConfig.php?Nav=".$NavHash."&kcid=".$KitchConfigID."' method='POST'>
  
  <div class='row'>

    <div class='col-sm-12'>
      <div class='md-form mb-0 form-sm'>
        <input value='".$KitchConfig."' type='text' id='configname".$KitchConfigID."' name='configname".$KitchConfigID ."' class='form-control'>
        <label for='configname".$KitchConfigID ."' class=''>Config Name:</label>
      </div>
    </div>

  </div>

  <div class='row'>
    <div class='col-sm-12'>
      <div class='md-form mb-0 form-sm'>
        <select id='designer_".$KitchConfigID."' name='designer_".$KitchConfigID."' class='mdb-select md-form' searchable='Search here..' onchange='changeInput(this.value, ".$SelField.");'>");
          $designersql = "SELECT * FROM `tbl_contacts` ORDER BY `FirstName`";
          $resultdesigner = $conn->query($designersql);

          if ($resultdesigner->num_rows > 0) {
          $resultdesigner->MoveFirst;
          echo("<option value='designer_".$KitchConfigID."'></option>");
          while($rowdesigner = $resultdesigner->fetch_assoc()) {
          $DesignerName = $rowdesigner["FirstName"]." ".$rowdesigner["LastName"];
          $DesignerID = $rowdesigner["ContactID"];
          if($rowdesigner["ContactID"]==$rowkitch["Designer"]){
          echo("<option value='".$DesignerID."' selected='selected'>".$DesignerName."</option>");
          }
          else {
          echo("<option value='".$DesignerID."'>".$DesignerName."</option>");
          }
          }
          }
          echo("</select>
        <input type='text' name='".$SelField."' value='' />
        <label for='designer_".$KitchConfigID."' class=''>Designer:</label>
      </div>
    </div>
  </div>
  
  
  <div class='md-form mb-0 float-right'>
    <button type='submit' class='btn btn-outline-danger waves-effect btn-sm float-right'>Save <i class='fas fa-magic ml-1'></i></button>
  </div>

For the Output i use:
 

    $upd_KitchConfigID = $_GET["kcid"];
    if(isset($_POST["configname$upd_KitchConfigID"]))
        {
           $upd_KitchConfig = $_POST["configname$upd_KitchConfigID"];
        }
        else
        {
            $upd_KitchConfig = 'NotSet';
            $ErrMsg = $ErrMsg . 'No KitchConfig, ';
        }
    
    if(isset($_POST["designer$upd_KitchConfigID"]))
        {
           $upd_KitchConfigDesigner = $_POST["designer_$upd_KitchConfigID"];
        }
        else
        {
            $upd_KitchConfigDesigner = 1;
            $ErrMsg = $ErrMsg . 'No Designer, ';
        }

echo("designer_$upd_KitchConfigID: ".$_POST["designer_$upd_KitchConfigID"]."<br><br>");

The value of ConfigName is well....BUT
for Designer i receive "Undefined index: designer..."

and i do not know why.
I'm shure it's a mistake by my stupid coding...but i can't find the error.

Maybe anybody of you guys can help me out?
thnx in advance!!!

BR,
Max

I want to make a table for each paddler_id (Field) which exists into pushup Database

My code for paddler_id = 1 is


require "../sql.php";

$result = mysql_query("EXPLAIN pushup");
$r = mysql_query("SELECT * FROM paddlerinfo t1 LEFT JOIN pushup t2 on t2.paddler_id=t1.id LEFT JOIN practicedate t3 on t3.practice=t2.practice_id ORDER BY t1.firstname ASC ");
$r1 = mysql_query("SELECT * FROM pushup  where paddler_id =1 ORDER BY practice_id ASC ");

echo "<table width='30%' border='1' cellpadding='0' cellspacing='0' style='font-family: monospace'>";
echo "<tr>";
echo "<td>DATE</td>";
while ($row = mysql_fetch_array($result)) {

if (in_array($row["Field"], array("paddler_id", "practice_id", "p_id"))) continue;

echo "<td>",($row["Field"]), "</td>";
}
echo "</tr>";

while (($data = mysql_fetch_array($r1, MYSQL_ASSOC)) !== FALSE) {
   unset($data["p_id"],$data["paddler_id"]);
echo "<tr>";
   foreach ($data as $k => $v) { 
echo "<td>";
      echo"$v";
echo "</td>";
}

echo "</tr>";
}


echo "</table>";

mysql_free_result($result);
mysql_free_result($r);
mysql_free_result($r1);
mysql_close();



The problem is on line $r1 = mysql_query("SELECT * FROM pushup  where paddler_id =1 ORDER BY practice_id ASC ");
I want to put something which makes it paddler_id = X where x = 2,3,4,5... and x exists in database and do not print twice the x (because that database may have rows where x = same id


How do i do that??

hello
in the attached code, how do i add another field to the query? ie, i have all the code to create the result and would like to add further values such as $dept? what is the correct way to code query? i must stress that i am using php 4.4.7 so json_encode is out. the code is working but just need to find a way to add the $dept to the qeury? many thanks

$dept  = array();
$box = array();

while ($row = mysql_fetch_array($result)) {
          $dept[] = $row['department'];
          $box[] = $row['custref'];
}

/*$items = rtrim($_POST['items'],",");
$sql = "UPDATE `boxes` SET status = 'Out'  WHERE Id IN ($items)";
$result = runSQL($sql);*/

$total = count(explode(",",$items)); 
$result = runSQL($sql);
$total = mysql_affected_rows(); 



/// Line 18/19 commented for demo purposes. The MySQL query is not executed in this case. When line 18 and 19 are uncommented, the MySQL query will be executed. 
header("Expires: Mon, 26 Jul 1997 05:00:00 GMT" );
header("Last-Modified: " . gmdate( "D, d M Y H:i:s" ) . "GMT" );
header("Cache-Control: no-cache, must-revalidate" );
header("Pragma: no-cache" );
header("Content-type: text/x-json");
$json = "";
$json .= "{\n";
$json .= "box: [\"". implode('","', $box) ."\"]\n";
$json .= "}\n";
echo $json;

$sql = "INSERT INTO `act` (`item`) VALUES ('". implode("'),('", $box) . "')";
$result = runSQL($sql);

Hi,

I am new to the boards and php and mysql.

I have created a database and can add entries via a form. I can query the database with another form and get the results to display in a table.

All good so far as that is what I was hoping to achieve but one of the fields I want to display as a hyperlink but I am having problems with the syntax as I keep getting errors when I try to wrap the variables in  <a href > tag.

Code: [Select]
<tr>
<td><?php echo $fquery['prefix']; echo $fquery['website']; ?><?php echo $fquery['website']; ?></td>
</tr>


now that displays e.g. http://www.example.comwww.example.com.

so I feel I am nearly there I just need advice as to how to construct a hyperlink from the fields queried.

ok, this is clearly 1st grade code for some, but i'm not there yet - I'm querying posts in WordPress, and the post_content will always have an image in the beginning of the post followed by the content. i don't want to get the image, just the content that's after the image, which is wrapped in anchor tags, of course.

Code: [Select]
<a href="http://path/to/image.jpg"><img src="http://path/to/image.jpg" /></a> <p>Post content yadda, yadda, hoowie</p>


obviously a character count won't work, so i need to get anything that follows the first "</a>", say...? is this the best way, or is there an easier way?

thanks for anyone's help.

GN

Hi there.  I'm totally new (about a week!) with php and mysql and am encountering a problem that perhaps someone can help me with? I've looked through to see if a similar problem has appeared or been solved, but without success, so apologies if I am repeating something.    In php I am trying to update 7 fields from a form from which a user has edited/modified any of the fields in a chosen record (except id).   Here is the code:   $id=$_GET['id'];