PHP - Merch List In Table With Limited Merch Per Table Row

Hi? im just a beginner in php i just want to ask how to insert a data into a table from a dropdown list. I have concatenate the itemid and description to form the dropdown list. But when i viewed my item_table the itemid and description columns are null. can you help me with this..
this is my php code for the dropdown list...
         <?php
         $query = "SELECT CONCAT(itemid,' ', '-',' ', description) AS Item FROM item_table";
         
         $result = mysql_query($query) or die(mysql_error());
         
         $dropdown = "<SELECT CONCAT(itemid,' ' '-',' ', description) AS Item FROM item_table>";

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

         $dropdown .= "\r\n<option value='{$row['Item']}'>{$row['Item']}</option>";

         }
         $dropdown .= "\r\n</select>";

         echo $dropdown;
         ?>

this my code for inserting data into the item_table...

<?php
if(isset($_POST ['submit']))
{
   $itemid = $_POST['itemid'];
   $description = $_POST['description'];
   $datein  = $_POST['datein'];
   $qtyin  = $_POST['qtyin'];
   $unitprice  = $_POST['unitprice'];
   $unit  = $_POST['unit'];
   $category  = $_POST['category'];
   $empid = $_POST['empid'];   
   $message ='';   
   if(($itemid && $description == "")||($itemid && $description == null))
   {
         header("location:IncomingEntry.php?msg=Incorrect");
         exit();
   }
   else
   {
      $link = mysql_connect('localhost', 'root', '') or die(mysql_error());
      $db_selected = mysql_select_db('inventory', $link);
      $message='';      
      $query = "INSERT INTO incoming_table (itemid , description,  datein,  qtyin, unitprice, unit, category, empid) VALUES ('".$itemid."',  '".$description."', '".$datein."',  '".$qtyin."', '".$unitprice."', '".$unit."', '".$category."', '".$empid."')";
      
      if (!mysql_query($query,$link))
      {
        die('Error: ' . mysql_error());
      }
      header("location: IncomingEntry.php?msg=1 record added");   
   }
}
?>

 

Hello everyone !!!

I am sorry if my vocabulary is not exact because english is not my first language. Also i am a newbie at PHP. I am doing this project for myself and if it work might be able to use it at work. But i am doing this to learn. I have been stuck on this problem for 2 weeks and i can t figure it out on my own. Ihave spend many hours searching forums but no success. Oh yeah i almost forgot some part of code are from me, some are scripts from internet i adapted.

I have a form with a dropdown menu and when i submit the form the value selected in the dropdown would be inserted in a table. The problem i have is that whatever the value i select it always inserts the last value of the dropdown in the table???

The form is made with the dropdown as an include. It is populated with values from an another table.
Here is the code for the dropdown list:
Code: [Select]
<form>
   <select name="nom_pcu_form" method="post">
<?php
$SQL = "SELECT * FROM pcu ORDER BY nom_pcu";
$res = mysql_query($SQL);
while($val=mysql_fetch_array($res)) 
{
$nom=$val["nom_pcu"];
$prenom=$val["prenom_pcu"];
$nom_complet = $prenom . $nom;
echo "<option>".$val["nom_pcu"].", ".$val["prenom_pcu"]."</option>\n";

$nom_pcu_form="".$val["nom_pcu"].", ".$val["prenom_pcu"].""; 
}
?>
</select>
</form>

Here is the part of the form wich calls the dropdown:

Code: [Select]
<form name="form2" method="post" action="Grille ecoute Permanent.php">
             
                <p>Nom :
                 <?php include 'liste_deroulante_pcu.php' ;?>
<p>no carte appel
                <input name="no_carte_appel" type="text" id="no_carte_appel">
                </p>
<?php  echo date("Y/m/d"); ?>

</form>

And this is the part where it is inserted in the table:

Code: [Select]
<?php
  if($_POST['doSubmit'] == 'Create')


  
  
  
mysql_query("INSERT INTO grille_ecoute_pcu_permanent (`user_name`,`nom_pcu`,`no_carte_appel`,`question_1`,`question_2`,`question_3`,`question_4`,
`question_5`,`question_6`,`question_7`,`question_8`,`question_9`,`question_10`,`question_11`,`question_12`,`question_13`,`question_14`,`question_15`,
`question_16`,`question_17`,`question_18`,`question_19`,`question_20`,`question_21`,`question_22`,`question_23`,`question_24`,`question_25`,`question_26`,
`question_27`,`question_28`,`question_29`)
 VALUES ('$user_name','$nom_pcu_form','$no_carte_appel','$question_1','$question_2','$question_3','$question_4','$question_5','$question_6',
 '$question_7','$question_8','$question_9','$question_10','$question_11','$question_12','$question_13','$question_14','$question_15','$question_16','$question_17',
 '$question_18','$question_19','$question_20','$question_21','$question_22','$question_23','$question_24','$question_25','$question_26','$question_27','$question_28',
 '$question_29')
 ") or die(mysql_error()); 
Note:$user_name and all $question are inserted correctly in the table.
$nom_pcu_form is the dropdown and it only records the last value of the dropdown even if it s not the value selected.
$no_carte_appel are not recorded at all

thanks for your time

I am working on a project that uses a drop down list to chose the category when inserting new data into the database. What I want to do now is make the drop down list default to the chosen category on the list records page and the update page. I have read several tutorials, but they all say that I have to list the options and then select the default. But since it is possible to add and remove categories, this approch won't work. I need the code to chose the correct category on the fly.

There are two tables, one that has the category ID and category name.  The second table has the data and the catid which is referenced to the category id in the first table.

Code: [Select]
--
-- Table structure for table `categories`
--

DROP TABLE IF EXISTS `categories`;
CREATE TABLE IF NOT EXISTS `categories` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `categories` varchar(37) NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=40 ;

-- --------------------------------------------------------

--
-- Table structure for table `links`
--

DROP TABLE IF EXISTS `links`;
CREATE TABLE IF NOT EXISTS `links` (
  `id` int(4) NOT NULL AUTO_INCREMENT,
  `catid` int(11) DEFAULT NULL,
  `name` varchar(255) NOT NULL DEFAULT '',
  `url` varchar(255) NOT NULL DEFAULT '',
  `content` varchar(255) NOT NULL DEFAULT '',
  PRIMARY KEY (`id`),
  KEY `catid` (`catid`)
) ENGINE=MyISAM  DEFAULT CHARSET=latin1 AUTO_INCREMENT=35 ;


Then is the list records file, I have

Code: [Select]
<?php
include ("db.php");
include ("menu.php");


$result = mysql_query("SELECT categories FROM categories") or die(mysql_error()); 
while ($row = mysql_fetch_array($result)) { 
$categories=$row["categories"];
$options.= '<option value="'.$row['categories'].'">'.$row['categories'].'</option>';
};



$id = $_GET['id'];
$query="SELECT * FROM links ORDER BY catid ASC";
$result=mysql_query($query);

?>
<table width="65%" align="center" border="0" cellspacing="1" cellpadding="0">
<tr>
<td>
<table width="100%" border="1" cellspacing="0" cellpadding="3">
<tr>
<td colspan="7"><strong>List data from mysql </strong> </td>
</tr>

<tr>
<td align="center"><strong>Category ID</strong></td>
<td align="center"><strong>Category ID</strong></td>
<td align="center"><strong>Name</strong></td>
<td align="center"><strong>URL</strong></td>
<td align="center"><strong>Content</strong></td>
<td align="center"><strong>Update</strong></td>
<td align="center"><strong>Delete</strong></td>
</tr>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td>
<SELECT NAME=catid>
<OPTION>Categories</OPTION>
<?php echo $options; ?>
</SELECT>
</td>
<td><? echo $rows['catid']; ?></td>
<td><? echo $rows['name']; ?></td>
<td><a href="<? echo $rows['url']; ?>"><? echo $rows['url']; ?></a></td>
<td><? echo $rows['content']; ?></td>
<td align="center"><a href="update.php?id=<? echo $rows['id']; ?>">update</a></td>
<td align="center"><a href="delete.php?id=<? echo $rows['id']; ?>">delete</a></td>

</tr>
<?php
}
?>
</table>
</td>
</tr>
</table>
<?php
mysql_close();
?>

So, how do I get this code
Code: [Select]
$result = mysql_query("SELECT categories FROM categories") or die(mysql_error());
while ($row = mysql_fetch_array($result)) {
$categories=$row["categories"];
$options.= '<option value="'.$row['categories'].'">'.$row['categories'].'</option>';
};

<SELECT NAME=catid>
<OPTION>Categories</OPTION>
<?php echo $options; ?>
</SELECT>

to give me an output that will be

something like if catid exactly matches categories.id echo categories.categorie  ???

so far everything I have done produces either a default category of the last category,  the catid (which is a number), all of the categories (logical since catid will always be = id, or nothing.  How do I get just the category name? 

I will keep reading and try to figure this out, but any help would be greatly appreciated.

Thanks in advance

Hi All, I want to copy into a table values from another table that partially match a given value, case-insensitively. So far I do as follows but I wonder whether there is a quicker way.

$input_table=array('1'=>'toto','2'=>'tota','3'=>'hello','4'=>'TOTO','5'=>'toto');
$input_table_2 = array_map('strtolower', $input_table);
$value_to_look_for='Tot';
$value_to_look_for_2=strtolower($value_to_look_for);
$output_table=array();
foreach ($input_table_2 as $k=>$v)
  {
  if(false !== strpos($v, $value_to_look_for_2))
    {
    $output_table[]=$input_table[$k];
    }
}

One drawback is that $input_table_2 is 'foreached' whereas there might be no occurrences, which would lead to a loss of time/resources for big arrays.   Thanks.

Hello,

I need some help. Say that I have a list in my MySQL database that contains elements "A", "S", "C", "D" etc... Now, I want to generate an html table where these elements should be distributed in a random and unique way while leaving some entries of the table empty, see the picture below. But, I have no clue how to do this... Any hints?



Thanks in advance,
Vero

Hi I am very new to PHP & Mysql. I am trying to insert values into two tables at the same time. One table will insert a single row and the other table will insert multiple records based on user insertion. Everything is working well, but in my second table, 1st Table ID simply insert one time and rest of the values are inserting from 2nd table itself. Now I want to insert the first table's ID Field value (auto-incrementing) to a specific column in the second table (only all last inserted rows).   Ripon.   Below is my Code:

<?php



$con = mysql_connect("localhost","root","aaa");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }


mysql_select_db("ccc", $con);


$PI_No = $_POST['PI_No'];
$PO_No = $_POST['PO_No'];


$qry = "INSERT INTO wm_order_entry (
Order_No,
PI_No,
PO_No) 


VALUES(
NULL,
  '$PI_No',
 '$PO_No')";
$result = @mysql_query($qry);




$val1=$_POST['Size'];
$val2=$_POST['Style'];
$val3=$_POST['Colour'];
$val4=$_POST['Season_Code'];
$val5=$_POST['Dept'];
$val6=$_POST['Sub_Item'];
$val7=$_POST['Item_Desc'];
$val8=$_POST['UPC'];
$val9=$_POST['Qty'];
$N = count($val1);
for($i=0; $i < $N; $i++)
{


$profile_query = "INSERT INTO order_entry(Size,
Style,
Colour,
Season_Code,
Dept,
Sub_Item,
Item_Desc,
UPC,
Qty,
Order_No
) VALUES( '$val1[$i]','$val2[$i]','$val3[$i]','$val4[$i]','$val5[$i]','$val6[$i]','$val7[$i]','$val8[$i]','$val9[$i]',LAST_INSERT_ID())";
$t_query=mysql_query($profile_query);
}
header("location: WMView.php");
mysql_close($con);
?>

Output is attached.

Hi All ,   I have a small table with 4 fields namely Day_ID, Dues, Last_Visit, Points. where Day_ID is an auto-increment field. The table would be as follows:   Day_ID -- Dues ---  Last_Visit ---  Points. 1  --------- 900 -------- 1/12  --------   6      2  --------- 700 -------- 4/12  --------   7 3  --------- 600 -------- 7/12  --------   5 4  --------- 600 -------- 9/12  --------   6 5  --------- 600 -------- 10/12 -------   6 6  --------- 600 -------- 14/12 -------   6   So this is the record of a person's visit to say a club. The last row indicates the last date of his visit to the club. His points on this date are 6. Based on this point value of 6 in the last row I want to retrieve all the previous BUT adjoining all records that have the same Points i.e. 6.    So my query should retrieve for me, based on the column value of Points of the last row (i.e. Day_ID - 6 ), as follows:   4  --------- 600 -------- 9/12  --------   6 5  --------- 600 -------- 10/12 -------   6 6  --------- 600 -------- 14/12 -------   6   This problem stated above had been completely resolved, thanks to a lot of help from Guru Barand by this following query :-

$query = "SELECT cv.day_id, cv.dues, cv.last_visit, cv.points
FROM clubvisit cv
WHERE last_visit >=
(
SELECT MAX(last_visit) FROM clubvisit
WHERE points <>
(
SELECT points as lastpoints
FROM clubvisit
JOIN
(
SELECT MAX(last_visit) as last_visit
FROM clubvisit
) as latest USING (last_visit)
)
)";

I am using this and it works perfectly except that now there is a slight change in the table because the criteria for points is now dependent on more than one column cv.points and is more like cv.points1, cv.points2, cv.points3 etc. So now I need to make a selection based on each of these cv.points columns.   As of now I can still get the results by running the query multiple times for each of the cv.points columns ( seperately for cv.points1, cv.points2, cv.points3) and it works correctly. However I am wondering if there is a better way to do this in just one go. This not only makes the code repetitive but also since the queries are interconnected, involves the use of transactions which I wish to avoid if possible.    The values that I require for each of the cv.point columns is  1. day_id of the previous / old day on which the cv.points value changed from the current day value, and  2. cv.points on that old/ previous day.    So for  example if the table is as below:   Day_ID -- Dues ---  Last_Visit ---  Points1 --- Points2. 1  --------- 900 -------- 1/12  -----------   9  ------------ 5    2  --------- 600 -------- 4/12  -----------   6  ------------ 6 3  --------- 400 -------- 7/12  -----------   4  ------------ 7 4  --------- 500 -------- 9/12  -----------   5  ------------ 8 5  --------- 600 -------- 10/12 ----------   6  ------------ 8 6  --------- 600 -------- 11/12 ----------   6  ------------ 8 7  --------- 600 -------- 13/12 ----------   6  ------------ 7 8  --------- 500 -------- 15/12 ----------   5  ------------ 7 9  --------- 500 -------- 19/12 ----------   5  ------------ 7   Then I need the following set of values : 1. day_id1 -- Day 7,      points1 ---- 6,          days_diff1 -- (9-7 = 2)  . // Difference between the latest day and day_id1 2. day_id2 -- Day 6,      points2 ---- 8,          days_diff2  -- (9-6 = 3) 3. day_id3 -- ....   and so on for other points.    Thanks all !

Hello everyone.

I need help with the following PHP APP. I am running on (Version PHP 7.2.10) I am trying to have a page table form on table.php pass the input variable of “5-Numbers” to another page called table_results.php

I want that variable string of “5-Numbers” to be compared against 4 arrays and output any duplicates found within each of those 4 lists. If nothing is found, I still want some visual output that reads “None found”.

 

Lets pretend I have the following example ..

On table.php, I typed inside my table form the 5-Numbers ..

INPUT:  2,15,37,13,28

On table_results.php, the 4 arrays to be compared against my input numbers “ 2,15,37,13,28” are ..

 

$array_A = array(2,6,8,11,14,18,24); 

$array_B = array(1,2,9,10,13,14,25,28,);  

$array_C = array(1,3,7,9,13,15,20,21,24); 

$array_D = array(4,5,12,22,23,27,28,29); 

 

So my output should read as follows below ..

OUTPUT: 

 

TABLE

COLUMN 1   COLUMN 2

ROW 1 Matches Found Results ..

ROW 2 GROUP A:            2 

ROW 3 GROUP B:             2,13,28

ROW 4 GROUP ?            13,15

ROW 5 GROUP ?            28

ROW 6 5#s Input:              2,15,37,13,28

 

Please let me know if anyone has any suggestions on how to go about it. 

Thanks.

Edited January 1, 2019 by Jayfromsandiego
Wanted to include image example

hi...

i have a table ...

i add and remove data in the table...when i add new record , information add to center of the table ! whats problem?   i want add data in first of table.   please guide me.thanks

 

 

I would appreciate your assistance, there are tons of login scripts and they work just fine. However I need my operators to login and then list their activities for the other operators who are logged in to see and if desired send their clients on the desired activity. I have the login working like a charm and the activities are listed just beautifully. How do I combine the two tables in the MySQL with PHP so the operator Logged in can only make changes to his listing but see the others.

FIRST THE ONE script the member logges in here to the one table in MSQL:
<?php
   session_start();
   
   require_once('config.php');
   
   $errmsg_arr = array();
   
   $errflag = false;
   
   $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);
   if(!$link) {
      die('Failed to connect to server: ' . mysql_error());
   }
   
   $db = mysql_select_db(DB_DATABASE);
   if(!$db) {
      die("Unable to select database");
   }
   
   function clean($str) {
      $str = @trim($str);
      if(get_magic_quotes_gpc()) {
         $str = stripslashes($str);
      }
      return mysql_real_escape_string($str);
   }
   
   $login = clean($_POST['login']);
   $password = clean($_POST['password']);
   
   if($login == '') {
      $errmsg_arr[] = 'Login ID missing';
      $errflag = true;
   }
   if($password == '') {
      $errmsg_arr[] = 'Password missing';
      $errflag = true;
   }
   
   if($errflag) {
      $_SESSION['ERRMSG_ARR'] = $errmsg_arr;
      session_write_close();
      header("location: login-form.php");
      exit();
   }
   
   $qry="SELECT * FROM members WHERE login='$login' AND passwd='".md5($_POST['password'])."'";
   $result=mysql_query($qry);
   
   if($result) {
      if(mysql_num_rows($result) == 1) {
         session_regenerate_id();
         $member = mysql_fetch_assoc($result);
         $_SESSION['SESS_MEMBER_ID'] = $member['member_id'];
         $_SESSION['SESS_FIRST_NAME'] = $member['firstname'];
         $_SESSION['SESS_LAST_NAME'] = $member['lastname'];
         session_write_close();
         header("location: member-index.php");
         exit();
      }else {
         header("location: login-failed.php");
         exit();
      }
   }else {
      die("Query failed");
   }
?>

................................................. ................................
Now I need the person who logged in to the table above to be able to make multiple entries to the table below

<?
$ID=$_POST['ID'];
$title=$_POST['title'];
$cost=$_POST['cost'];
$activity=$_POST['activity'];
$ayear=$_POST['aday'];
$aday=$_POST['ayear'];
$seats=$_POST['special'];
$special=$_POST['seats'];
mysql_connect("xxxxxx", "xxx350234427", "========") or die(mysql_error());
mysql_select_db("xxxx") or die(mysql_error());
mysql_query("INSERT INTO `activity` VALUES ('ID','$title', '$cost','$activity', '$aday', '$ayear', '$special', '$seats')");
Print "Your information has been successfully added to the database!"
?>
Click <a href="member-profile.php">HERE</a> to return to the main menu
  <?php      
?>

Not sure how to do this at all... I'm creating a page with a form to edit existing info in two tables in the database. I need to pre-check some checkboxes and I don't know how. The first query query_selectguest below puts data into the form just fine. query_checked looks in the second table for any rows that match the discount id. Then further down in the form, I have a php snippet that pulls all the classes from a third database. As I'm echoing these out to the page, I want them to be pre-checked if they match a result found in $result_checked

here's what I have right now...

Code: [Select]
$query_selectguest = 'SELECT * FROM tbl_discount WHERE discount_id='.$passedID;
$result_guest = mysql_query($query_selectguest);
$g_row = mysql_fetch_array($result_guest);

$query_checked = 'SELECT * FROM active_discounts WHERE disc_id='.$passedID;
$result_checked = mysql_query($query_checked);
$h_row = mysql_fetch_array($result_checked);

?>

<form name="form1" method="post" action="update_discount.php" onSubmit="return validate_form()">

<input name="discount_name" type="text" class="input" id="subject" size="50" maxlength="100" value="<? print $g_row['discount_name'] ?>">
<input name="discount_amount" type="text" class="input" id="subject" size="5" maxlength="3" value="<? print $g_row['discount_amount'] ?>">

<!-- here's what I'm concerned with -->

This discount applies to these classes:<br>
         
           <?php
$quer4=mysql_query("SELECT workshop_id, workshop_title FROM tbl_workshops order by workshop_title");

while($row4 = mysql_fetch_array($quer4)) {
echo "<input type='checkbox' name='workshop_link_1[]' value='".$row4[workshop_id]."'>".$row4[workshop_title]."<BR>";
}

 ?>

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=317025.0

I know I'm doing it something right, but can someone tell me why only one table is showing up? Can you help me fix the issue?

Heres my code:


function showcoords()
{
echo"J3st3r's CoordVision";
$result=dbquery("SELECT alliance, region, coordx, coordy FROM ".DB_COORDFUSION."");
dbarray($result);
$fields_num = mysql_num_fields($result);

echo "<table border='1'>";
// printing table headers


    echo "<td>Alliance</td>";
    echo "<td>Region</td>";
    echo "<td>Coord</td>";


// printing table rows
while($row = mysql_fetch_array($result))
{
    

    // $row is array... foreach( .. ) puts every element
    // of $row to $cell variable
    foreach($row AS $Cell)
     echo "<tr>";     
     echo "<td>".$row['alliance']."</td>\n";
     echo "<td>".$row['region']."</td>\n";
     echo "<td>".$row['coordx'].",".$row['coordy']."</td>\n";
     echo "</tr>\n";
    
}
echo "</table>";
mysql_free_result($result);

}



I have 2 rows inserted into my coords table. Just frustrated and ignorant to php.

I am trying to print the list of a table which I requested with "SELECT DISTINCT" as below

Code: [Select]
$db_connect = mysqli_connect(DB_HOST, DB_USER, DB_PASSWORD, DB_NAME);

$sql_get = "SELECT DISTINCT category FROM con";

$sql_run = mysqli_query($db_connect, $sql_get) or mysqli_error($db_connect);

$sql_assoc = mysqli_fetch_assoc($sql_run);

What is now needed to print the list of the table data by this conditions?

I tried the while loop, but I seem to approach wrong, and get endless loops or errors.

I'm setting up a newsletter thing for my website.

I have a newsletter table in MySQL:
Quote

+----------------+------+--------+
| email          | mens | womens |
+----------------+------+--------+
| test2@test.com |    1 |      1 |
| test1@test.com |    1 |      0 |
+----------------+------+--------+


I am using a HTML form and this PHP code I learn't from the manual, which sends out e-mail's.
PHP:
if ($_POST['newsletter'] == 'Mens') {
   $to = '';
   $subject = $_POST['subject'];
   $body = $_POST['body'];
   $header = 'From: Me Someone <me@someone.com>';
   mail($to, $subject, $body, $header);
}

What I want to do with the above code is send out an e-mail to all the e-mails in my MySQL database that are tagged '1' under mens.

How would I go about doing this? I'm guessing I will have to use a MySQL query in the $to = ''; that goes something like this:
$to = '$query (select from `newsletters` where `email` = 1');
?