PHP loop within a loop, option selected?

Question 1

Question by David

Using PHP I echo out table rows in a loop like this:

<?php
/* SQL STUFF */

  while ($row = mysql_fetch_array($select_courseelements)) {
   echo "<tr>n";
    echo "<td>".$row['scpe_name']."</td>n";
    echo "<td>".$row['scpe_days']."</td>n";
   echo "</tr>n";
  }

Now I would like to include a <select> element with 5 predefined <option> values inside a <td> running with the loop. The option values will be 1 to 5.

There is also a column inside the $row loop that holds a value of 1 to 5 ($row['scpe_grades_status']).

Each time this value is equal to the one in the <select> I want it to change it to selected='selected'.

Would this be possible?

My <select> will look something like this when it’s beeing run in the loop:

echo "<td>n";
echo "<select id='elements_grade'>n";
        echo "<option value='1'>Registrerad</option>n";
        echo "<option value='2'>Ej påbörjad</option>n";
        echo "<option value='3'>Pågående</option>n";
        echo "<option value='4'>Godkänd</option>n";
        echo "<option value='5'>Deltagit</option>n";
        echo "<option value='6'>Ej deltagit</option>n";
echo "</select>n";
echo "</td>n";

Answer

Answer by piers

piers

$array = array('Registrerad' => 1, 'Ej påbörjad' => 2, 'Pågående' => 3, 'Godkänd' => 4, 'Deltagit' => 5, 'Ej deltagit' => 6);

foreach ($array as $key=>$value) {
    if ($value == $row['scpe_grades_status'])
        echo '<option value="'.$value.'" selected>'.$key.'</option>';
    else
        echo '<option value="'.$value.'">'.$key.'</option>';
}

Something like that?

Answer

Answer by Starx

Starx

Sure, build the values from a loop. and you can compare the values from that part.

for($i = 1; $i<=5; $i++) {
   echo "<option value='$i'";
   echo ($row['scpe_grades_status'] == $i) ? " selected='selected'": "";
   echo ">...."</option>"
}

Question 2

Question by Baruch

Baruch

I am trying to create a select drop-down list using php. Every time i try, i get an error.
Here is my code:

The function:

function dropDown(){


$options="<select>"; 
$connect = mysql_connect('localhost','id','pass') or die ("couldn't   connect!").mysql_error; 

mysql_select_db('db') or die('could not connect to database!');


$sql="SELECT * FROM DESC"; 
$result=mysql_query($sql); 


while ($row=mysql_fetch_array($result)) {  ****this is line 60

$name=$row["name"]; 

$options.="<option value="$name">".$name."</option>"; 
} 

$options.= "</SELECT>";
return "$options";
}

and then i just call it in my code

<?php  
 include ('includes/functions.php');
....



$list = dropDown();
echo "$list";


....
>

the error i get is:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/garagenj/public_html/dispatch/includes/functions.php on line 60

Answer

Answer by JT Smith

JT Smith

If your table name is DESC then that’s your problem. It’s all about namespacing. You can’t even have a field named desc, I’ve tried. It errors everytime

Answer

Answer by Starx

Starx

SELECT * FROM DESC ??????

From what table
Or, if the table name is DESC. escape them with ticks

And You dont have start a new connection, every time you call the function. Put your connection part somewhere else

$connect = mysql_connect('localhost','id','pass') or die ("couldn't   connect!").mysql_error; 
mysql_select_db('db') or die('could not connect to database!');

function dropDown(){
    $options="<select>"; 
    $sql="SELECT * FROM `DESC`"; 
    $result=mysql_query($sql); 

    while ($row=mysql_fetch_array($result)) { 

        $name=$row["name"]; 

        $options.="<option value="$name">".$name."</option>"; 
    } 

    $options.= "</SELECT>";
    return $options;

}

Question 3

Question by Malyo

Malyo

I’m looking for best solution to this problem. I have a simple shop logical problem. There are 2 select elements, size and color. I want to make them dependent, on data (now it’s example data, but later it’s gonna be from database) – size will decide which color options will be visible for customer (hiding not necessary ones).

First problem is that when i make change event, and i wanna hide the default shown element on document ready, it’s still visible (i’d have to change color to different than open dropdown again and it won’t be visible then).

Second is that i’m looking for most flexible solution, since i have doubts about mine. Here’s the code:

       var rozmiar = new Array("S", "M", "L", "XL", "XXL");
   var kolor = new Array("Czerwony", "Niebieski", "Zielony", "Biały", "Czarny");
   var opcje = new Array( rozmiar, kolor);

        $(document).ready(function(){
        $('.form1').change(function(){
                $('.form2 option').show();

                var selectSelector = function(z){
                    selectSelector = $('select.form2 option[value='+kolor[z]+']').hide();
                };

                wybranyRozmiar = $(this).val();
                    if(wybranyRozmiar == rozmiar[0]){
                        selectSelector(0);
                    }
                    if(wybranyRozmiar == rozmiar[1]){
                        selectSelector(1);
                    }
                    if(wybranyRozmiar == rozmiar[2]){
                        selectSelector(2);
                    }
                    if(wybranyRozmiar == rozmiar[3]){
                        selectSelector(3);
                    }
                    if(wybranyRozmiar == rozmiar[4]){
                        selectSelector(4);
                    }
            });
        });

Answer

Answer by Starx

Starx

I am answering the only part I understand.

Instead of using multiple if statements you can use switch

switch(selectsize) {
   case rozmiar[1]:
       $('select.form2 option[value='+color[2]+']').hide();
       break;
   //case <another>"
       //break;
}

Tag: option

PHP loop within a loop, option selected?

Question by David

Answer by piers

Answer by Starx

SELECT * FROM DESC ??????

Jquery form onchange select values changing

Question by Malyo

Answer by Starx

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