PHP - Insert Form Variable In First Table And While Loop Variables (array) In The 2nd
First page adds a new job number, then the order page loaded with the job number id as a get id. Basically there is a while loop in the order page which shows products/services client can order and he chooses what he requires 'one or eight services' (8 in total) and some other variables like date of order and client name etc. Here is the order item code
The first sql statement executes fine, but in the second sql query nothing happens $sql="insert into job_order(order_num,order_date,order_customer_id, order_remarks) values(".$_GET['id'].",NOW(),".$_POST['companyBox'].",'".$_POST['remarkBox']."');"; $res=mysql_query($sql); $id=mysql_insert_id(); foreach($_POST as $key => $value) { if(!empty($value)) { $key.' => '.trim(strip_tags($value)); $order="INSERT INTO orderprod (order_num,prod_id,order_amount,teeth_amount) VALUES ('$_GET[id]','$value','$value','$value');"; $orderres=mysql_query($order) or die(mysql_error()); } }?>$sql="insert into job_order(order_num,order_date,order_customer_id, order_remarks) values(".$_GET['id'].",NOW(),".$_POST['companyBox'].",'".$_POST['remarkBox']."');"; $res=mysql_query($sql); $id=mysql_insert_id(); foreach($_POST as $key => $value) { if(!empty($value)) { $key.' => '.trim(strip_tags($value)); $order="INSERT INTO orderprod (order_num,prod_id,order_amount,teeth_amount) VALUES ('$_GET[id]','$value','$value','$value');"; $orderres=mysql_query($order) or die(mysql_error()); } } ?> ============= This is the formI have removed parts which are irrelevant. Please note that dateBox and companyBox are not required to be looping as they are only for first table, echo "<form action=".$config_basedir."./vieworder.php?id=".$_GET['id']." name=form1 method=post>";?><table><tr><td><h4>JOB ORDER</H4></TD><TD></TD></TR><TR><TD>ORDER NUMBER</TD><TD><?PHP ECHO $_GET['id'] ?></td></tr><tr><td>ORDER DATE</td><td><input type=text name=dateBox></td></tr> <tr><td>COMPANY NAME</td><td><?PHP$sql="select * from customers";$res=mysql_query($sql);echo "<select name=companyBox><option value=''>Please select</option>";WHILE($fetch=mysql_fetch_assoc($res)){ echo "<option value='".$fetch['id']."'>".$fetch['cust_name']."</option>";} echo "</select>"; echo "</td></tr>"; ?> </table> <table><tr><th>ITEM</th><th>QUANTITY</th><th>N0. of Teeths</th></tr><tr><?PHP$sql="select * from products";$res=mysql_query($sql); WHILE($fetch=mysql_fetch_assoc($res)){ echo "<td><input type=text name=desBox value='".$fetch['prod_id']."'>".$fetch['prod_name']."</td><td><input type=text name=quantBox></td><td><input type=text name=teethBox></td>";echo "</tr>";}echo "</table>";?>echo "<form action=".$config_basedir."./vieworder.php?id=".$_GET['id']." name=form1 method=post>"; ?> <table> <tr> <td><h4>JOB ORDER</H4></TD><TD></TD> </TR> <TR> <TD>ORDER NUMBER</TD><TD><?PHP ECHO $_GET['id'] ?></td> </tr> <tr> <td>ORDER DATE</td><td><input type=text name=dateBox></td> </tr> <tr> <td>COMPANY NAME</td><td> <?PHP $sql="select * from customers"; $res=mysql_query($sql); echo "<select name=companyBox><option value=''>Please select</option>"; WHILE($fetch=mysql_fetch_assoc($res)){ echo "<option value='".$fetch['id']."'>".$fetch['cust_name']."</option>";} echo "</select>"; echo "</td></tr>"; ?> </table> <table> <tr> <th>ITEM</th><th>QUANTITY</th><th>N0. of Teeths</th> </tr> <tr> <?PHP $sql="select * from products"; $res=mysql_query($sql); WHILE($fetch=mysql_fetch_assoc($res)){ echo "<td><input type=text name=desBox value='".$fetch['prod_id']."'>".$fetch['prod_name']."</td> <td><input type=text name=quantBox></td> <td><input type=text name=teethBox></td>"; echo "</tr>";} echo "</table>"; ?> HERE IS THE IMAGE showing the populated services. http://dubads.com/images/order.jpg Similar TutorialsGood evening,
I am working on a program that takes images, and casts them into an associated array, I then want to display thumbnails of those images into a 3 column by 4 row table...
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(For my visual friends)
here is my current, pardon for the mess, but right now, functionality is more important.
<!DOCTYPE html> <html> <head> <title>Zodiac Gallery</title> </head> <body> <h2 style="text-align:center">Zodiac Gallery</h2> <p>Click a thumbnail image to see enlarged view.</p> <?php $ZodiacArray = array ( "Images/rat.jpg" => "Rat", "Images/ox.jpg" => "Ox", "Images/tiger.jpg" => "Tiger", "Images/rabbit.jpg" => "Rabbit", "Images/dragon.jpg" => "Dragon", "Images/snake.jpg" => "Snake", "Images/horse.jpg" => "Horse", "Images/goat.jpg" => "Goat", "Images/monkey.jpg" => "Monkey", "Images/rooster.jpg" => "Rooster", "Images/dog.jpg" => "Dog", "Images/pig.jpg" => "Pig"); foreach ($ZodiacArray as $image => $zodiac) { echo "<table> <tr> <td><img src="$image" alt="$zodiac" style="width:68px;height:65px"></td <td><img src="$image" alt="$zodiac" style="width:68px;height:65px"></td> <td><img src="$image" alt="$zodiac" style="width:68px;height:65px"></td> </tr> <tr> <td><img src="$image" alt="$zodiac" style="width:68px;height:65px"></td> <td><img src="$image" alt="$zodiac" style="width:68px;height:65px"></td> <td><img src="$image" alt="$zodiac" style="width:68px;height:65px"></td> </tr> <tr> <td><img src="$image" alt="$zodiac" style="width:68px;height:65px"></td> <td><img src="$image" alt="$zodiac" style="width:68px;height:65px"></td> <td><img src="$image" alt="$zodiac" style="width:68px;height:65px"></td> </tr> <tr> <td><img src="$image" alt="$zodiac" style="width:68px;height:65px"></td> <td><img src="$image" alt="$zodiac" style="width:68px;height:65px"></td> <td><img src="$image" alt="$zodiac" style="width:68px;height:65px"></td> </tr> </table>"; ?>Help is welcome, thank you. 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. Hello, If I have this string: $tags="baseball glove face" (this can vary from 1 to 15 different words) how would i loop through and divide these into different $tag1=baseball $tag2=glove $tag3=face .. and so on (if more words) Thanks for help I have this simple while loop which retrieves data from a mysql query and displays several links on my homepage. I would like to avoid using the php get function and add query strings to my urls I am thinking of using session variables but I need help and I'm pretty sure this can't be done. When a visitor clicks a link from the several ones displayed by the while loop, that particular variable would be set in a session. In my code, the session will always send the last var Can this be done? Code: [Select] <? session_start(); // Start Session Variables $result = mysql_query("my query"); while($slice = mysql_fetch_assoc($result)){ $url = $slice['url']; $name = $slice['name']; ?> <a href="<? echo $url; ?>"><? echo $name; ?></a> <? } $_SESSION['name'] = $name; // Store session data ?> I need a help in the following : I have an admin page from where the admin attaches an gif image.This attached image should be shown to the user as a scrolling image. The code for the scrolling image is done through javascript. So my actual problem is getting the name of the attached gif image to the javascript array which is used for scrolling horizontally. Hello All, Attached are 2 screenshots of the form and output for better reference. What I am trying to do is select all or some checkboxes and for those boxes that are checked, echo out their Payment Method. ---If a checkbox is checked but the Payment Method box == select, give an error. ---If checkbox is checked and Payment Method != select, echo out the ID and selection from the drop down box. Currently, I am echoing the correct ID(s) when certain boxes are checked, but I'm getting every drop down box selection when the corresponding box is not selected. Any help would be greatly appreciated. This is the HTML / PHP Form: $results .= '<table align="center" width="70%" border="0" bgcolor="'.$bg.'" cellpadding="2" cellspacing="0" class="resultsUnderline">'; $results .= '<tr>'; $results .= '<td width="30%" align="left">'.'<input type="checkbox" name="checkbox[]" value="'.$app_pmt_id.'" id="checkbox" />' .$full.'<FONT size="1" color="maroon"> ( '.$app_id.' )</FONT></td>'; $results .= '<td width="30%" align="center">'.$created.'</td>'; $results .= '<td width="20%" align="center"> <select name="pay_method[]" class="drop" id="pay_method"> <option value="select">-- Method --</option> <option value="check">Check</option> <option value="credit">Credit Card</option></td>'; $results .= '<td width="10%" align="center">'.$paid.'</td>'; $results .= '<td width="10%" align="right">'.$fee.'</td>'; $results .= '</tr>'; $results .= '</table>'; Here is the PHP getting the results: if(isset($_POST['action_box']) && $_POST['action_box'] == 'mark_paid') { $checked = array(); $checked2 = array(); if(!isset($_POST['checkbox'])) { $val_error[] = 'You have not selected any applications.'; } elseif($_POST['pay_method'] == 'select') { $val_error[] = 'Please select a Payment Method.'; } else { foreach($_POST['checkbox'] as $value) { $checked[] = $value; } foreach($_POST['pay_method'] as $value2) { $checked2[] = $value2; } echo implode(",", $checked); echo implode(",", $checked2); //$result = mysql_query("UPDATE app_pmt SET paid = '1' WHERE app_pmt_id IN(" . implode(",", $checked) . ")"); //header('Location: bal_due.php?id='.$_GET['id']); } } This is really blowing my mind... the code is as follows the table, the column names everything is right and contains information but the variables aren't being filled... I don't get it... Anyone have any ideas or options I might not have thought of for debugging? $item_name = ("SELECT * FROM items WHERE id = '$grab_item'"); $que_items = mysql_query($item_name); while ($item_todel = mysql_fetch_array($que_items, MYSQL_BOTH)) { $name = $item_todel['name']; $desc = $item_todel['desc']; } echo ($name . "<br>" . $desc . "<br>" . $grab_item); the only variable thats set is grab_item, fetch_array throws no errors, if I plug in the query straight to the sql I get all the proper information... It's just not filling the two variables.... Hello Guys, I want to insert an array variable into the database e.g Code: [Select] foreach($html->find('div[class=PostContent]') as $element) { echo $element; $sq = "INSERT INTO articles(original_text) VALUES ('$element') WHERE article_link='$item_url'"; $result = mysql_query($sql1) or die('Query failed: ' . mysql_error()); } i want to insert the variable $element into the database but i'm not able to do so for some reason! $element contains only a paragraph of text. how can i insert the variable $element? I am making a website where writers can share stories. One of the fields is other stories. So I used this code to make it a array with the links in each f the items. $other_split = explode(',', $other_form); $other_int = count($other_split); $x = 0; while ($x>$other_int) { $other_link[$x] = "<a href = '$other_split[$x]'> Chapter $x </a>"; } How can I insert the data in that array into one non array variable. Hello, I'm working on a PHP register form, all I want to do, is to be able to insert an extra variable into the database: Code: [Select] // now we insert it into the database $insert = " INSERT INTO users (username, password) VALUES ('".$_POST['username']."', '".$_POST['pass']."') "; $add_member = mysql_query($insert); right, so I have the username and password variable being passed. But I also want to pass a variable into a column called 'weaponAttachments', all I want for this variable to be is 5000. So, in other words, something like this: 5000['weaponAttachments'] Thanks I'm not versed in PHP OOP and I have some code that I need to echo out onto a page. I don't know what the meaning of $ct->something is in this code. Is this an array of some type? How do I echo out the $ct->title onto another page? Use a function call? I just don't know what all of the $ct variables are and how to get to them. Any help is appreciated. Code: [Select] function current_theme_info() { $themes = get_themes(); $current_theme = get_current_theme(); if ( ! isset( $themes[$current_theme] ) ) { delete_option( 'current_theme' ); $current_theme = get_current_theme(); } $ct->name = $current_theme; $ct->title = $themes[$current_theme]['Title']; $ct->version = $themes[$current_theme]['Version']; $ct->parent_theme = $themes[$current_theme]['Parent Theme']; $ct->template_dir = $themes[$current_theme]['Template Dir']; $ct->stylesheet_dir = $themes[$current_theme]['Stylesheet Dir']; $ct->template = $themes[$current_theme]['Template']; $ct->stylesheet = $themes[$current_theme]['Stylesheet']; $ct->screenshot = $themes[$current_theme]['Screenshot']; $ct->description = $themes[$current_theme]['Description']; $ct->author = $themes[$current_theme]['Author']; $ct->tags = $themes[$current_theme]['Tags']; $ct->theme_root = $themes[$current_theme]['Theme Root']; $ct->theme_root_uri = $themes[$current_theme]['Theme Root URI']; return $ct; } hi ..
hit little snag Ive got html <form name="formx" method="POST" action="connect.php" class='ajaxform'> and about 4 varibles in that form .. now how to properly
query table users so i can find out which prefix correspond with that user
Insert into table "upis" in this case that prefix with 3 variables..
everything in single,.. in this case connect.php file
I know how to do QUERY and INSERT INTO but i don't know how to connect with 2 different tbl or db and do the query and input of data.. every time I put 2 of them together .. never works out..
this is example.. i dont know is this right way to do it .. I am trying to insert records from an foreach array, it only inserts the first record and I cannot work out how to get it to loop through all of the records. Thank you foreach($_SESSION['post_data_array'] AS $seat) { $rowId = substr($seat, 0, 1); $columnId = substr($seat, 1); echo $rowId . $columnId . ", "; } $sql125 = "INSERT INTO booked_seats(booking_id, row_id, column_id) values ('$book_id', '$rowId', '$columnId')"; $result125 = mysql_query($sql125); if ($result125) { echo "worked"; } else { echo "didnt work"; } Hi guys
A mobile application send data in Json format, so I want to insert the data into Mysql table, when I try to insert I have gotten the error "PHP Warning: mysqli::prepare() expects exactly 1 parameter, 3 given", my table only has 2 columns, "id" and "name" column.
The array has this data
( [0] => ('id', '2') [1] => ('name', 'Jhon') )The function to parse and insert data ... function redeem() { // Check for required parameters $json = file_get_contents('php://input'); $obj = json_decode($json,true); // Assumes $obj == array(0 => $assocArray, 1 => $assocArray) foreach($obj as $index => $assocArray) { // Assumes $assocArray == array(0 => array('id' => '2'), 1 => array('name' => 'John')) $stmt = $this->db->prepare('INSERT INTO prueba (id,nombre) VALUES (%d,%s)',$assocArray[0]['id'],$assocArray[1]['name']) or die(mysqli_error($this->db)); $stmt->execute(); } } Edited by Maq, 06 June 2014 - 03:54 PM. Hello, I'm working on a form that once submitted, it inserts the result into my SQL table, but I'm trying to get it so it will update an imploded array. EG: current SQL table is "red,blue,green". Then user submits 'white', the SQL table should then update to: "red,blue,green,white" (notice how its added to the end). This is what I have so far: Code: [Select] $attack_out = $info['attack_out']; $attack_out_split = explode(",", $attack_out); if(isset($_POST["submit_attack"])){ $attack_user = $_GET["attack"]; // user - sent from form //$troops_send = $_POST["troop_send"]; // ammount - sent from form $attack_user_array = array(); $i = 0; $max = (count($attack_out_split) + 1); while($i < $max){ $reverse = ($max - 1) - $i; $attack_user_array[$i] = $attack_out_split[$reverse]; $attack_user_array[$max - 1] = $attack_user; $i++; } $attack_user_glue = implode(",",$attack_user_array); $result = mysql_query("UPDATE users SET attack_out='$attack_user_glue' WHERE username='$username'"); } I can't figure out the algorithm in my while loop. Would appreciate any help, thanks Hi again all, Why does the foreach loop im doing, put the array values from collection, into seperate table rows rather than 1 row per whole array? <?php class registration{ public $fields = array("username", "email", "password"); public $data = array(); public $table = "users"; public $dateTime = ""; public $datePos = 0; public $dateEntryName = "date"; public $connection; function timeStamp(){ return($this->dateTime = date("Y-m-d H:i:s")); } function insertRow($collection){ //HERE foreach($this->fields as $row => $value){ mysql_query("INSERT INTO $this->table ($value) VALUES ('$collection[$row]')"); } mysql_close($this->connection->connectData); } function validateFields(){ $this->connection = new connection(); $this->connection->connect(); foreach($this->fields as $key => $value){ array_push($this->data, $_POST[$this->fields[$key]]); } $this->dateTime = $this->timeStamp(); array_unshift($this->data, $this->dateTime); array_unshift($this->fields, $this->dateEntryName); foreach($this->data as $value){ echo "$value"; } $this->insertRow($this->data); } } $registration = new registration(); $registration->validateFields(); ?> I end up with 3 rows row 1: username row 2: email row 3 : password rather than row 1:username email password. I have an array of data in which I want to display in a table with the first row of data to not be formatted. In the second row of data, I want it formatted with the CSS class of "success". When I run this code, I do not get any formatting of any rows in the table. I have tried various different options without success. Any help would be greatly appreciated.
<?php Copied wrong set of code 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?? HI first timer here so please forgive me if this is in the wrong section. Anyway, here goes. I have 2 tables Table 1 Levels LevelID LevelName Table 2 Members MemberID FirstName LastName Birthdate LevelID I have a form that has the LevelID populated from table 1 with a query. When I enter all my data and hit submit the data gets written to the members table. However, the LevelID is written as 0 for each and everyone record that I have entered. So the question is what am I missing? As mentioned I am very new at this so any help would be greatly appreciated. Please see below for my code. Thanks in advance <?php require_once 'dblogin1.php'; $db_server = mysql_connect($db_hostname, $db_username, $db_password); if (!$db_server) die("Unable to connect to MySQL: " . mysql_error()); mysql_select_db($db_database) or die("Unable to select database: " . mysql_error()); $sql = "SELECT\n" . "levels.LevelID,\n" . "levels.LevelName\n" . "FROM\n" . "levels\n"; $result = mysql_query($sql) or die(mysql_error()); $options=""; while($row = mysql_fetch_array($result)){ $LevelId=$row["LevelID"]; $LevelName=$row["LevelName"]; $options.="<OPTION VALUE=\"$LevelID\">".$LevelName.'</option>'; } ?> <style type="text/css"> <!-- body p { color: #F00; } --> </style> <table width="500" border="1" align="center" cellpadding="0" cellspacing="1" > <tr> <td> <form name="form1" method="post" action="insert_ac.php"> <table width="100%" border="1" cellspacing="1" cellpadding="3"> <tr> <td colspan="3"><div align="center"> <p><strong>Myers Player Registration Form</strong></p> </div></td> </tr> <tr> <td>First Name</td> <td><div align="center">:</div></td> <td><input name="firstname" type="text" id="firstname"></td> </tr> <tr> <td>Last Name</td> <td><div align="center">:</div></td> <td><input name="lastname" type="text" id="lastname"></td> </tr> <tr> <td>Birthdate (yyyy-mm-dd)</td> <td><div align="center">:</div></td> <td><input name="birthdate" type="text" id="birthdate"> </td> </tr> <tr> <td>Level</td> <td><div align="center">:</div></td> <td><select name="LevelID"> <OPTION VALUE="0">Choose <?php echo $options?> </select></td> </tr> </form> </td> </tr> </table> |