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PHP - Two Foreach To Inset In Database
Hi, I am stuck on this point for few hours please can someone help.
From one html form i am submitting two input boxes as array Code: [Select] <input type="text" name="names[]" /> <input type="text" name="position[]" /> <input type="text" name="names[]" /> <input type="text" name="position[]" /> <input type="text" name="names[]" /> <input type="text" name="position[]" /> Now when i submit i need to insert that values in a relational table ID name position I tried this way but dont know if i need to run two loops. $names = $_POST('names'); $positions = $_POST('position'); // now here i dont know how i will insert in database foreach ($names as $name) { foreach ($positions as $position) { $mysqlquery = ("INSERT into TABLE (name,position) values ($name,$position)"); } } Please any help how i will insert that Thanks Similar TutorialsI am extracting data from one of the sites.I would join these sentences separated by , (comma). The solution that comes to my head is by inserting the text in the file and then using file functions.However, i was wondering if i can use str_replace here . Can anyone please point me in the right direction. Code: [Select] Hospital 30-Day Death (Mortality) Rates from Heart Attack: 15.2% (No different than U.S. National Rate) Hospital 30-Day Readmission Rates from Heart Attack: % (Number of Cases Too Small*) Hospital 30-Day Death (Mortality) Rates from Heart Failu 10.5% (No different than U.S. National Rate) Hospital 30-Day Readmission Rates from Heart Failu 29.9% (Worse than U.S. National Rate) Hospital 30-Day Death (Mortality) Rates from Pneumonia: 9.8% (No different than U.S. National Rate) Hospital 30-Day Readmission Rates from Pneumonia: 20.6% (No different than U.S. National Rate) I am stuck on inserting the data into mysql via php. Its a foreach method and I am just stuck Code: [Select] include '../Database/take_an_exam.php'; $intNum = 1; $intnumber = 1; while( $info = mysql_fetch_array( $sqll )){ echo "<input type='hidden' name=\"Que_ID\" value=\"{$info['Que_ID']}\" /> "; echo " $intNum, {$info['Que_Question']} <br />\n"; $intNum++; for ($i =1; $i < 5; $i++) { echo "<input type=\"checkbox\" name=\"choice[{$info['Que_ID']}.$i][]\" />{$info['Que_Choice'.$i]}<br />\n"; $intnumber++; } } ?> <input type="submit" value="submit" name="submit"> This is my form that can output the Question and its choices. I am echoing this on to the next page, the code is shown below Code: [Select] <?PHP session_start(); $name= $_SESSION['username1']; echo "User ID: $name <br/>"; $gender = $_POST["choice"]; $que_ID = $_POST["Que_ID"]; foreach ($gender as $key => $array) { if($array) { $val=1; } ///echo "Question ID and the choice ID:$key. Value is:.$val <br />\n"; $well = array(floor($key), substr(strstr($key, '.'), 1)); //$well = array(floor($key), $key - floor($key)); echo "Question ID: $well[0], "; echo "Choice: $well[1]. value: $val<br/>"; } //}break; ?> Now, instead of echoing. I just want to add this to mysql. The layout of my sql is Que_ID, Ans_Choice1, Ans_Choice2, Ans_Choice3, Ans_Choice4, Use_ID When I echo my php my outcome is Quote User ID: 2 Question ID= 1 Choice 1 Question ID= 2 Choice 3 Question ID= 3 Choice 4 Question ID= 4 Choice 1 Hello everyone, Let's say I am connecting to a database using PDO (I am excluding prepared statements for the sake of simplicity): try { $conn = new PDO('mysql:host=localhost;dbname=cms', 'root', 'password'); $result = $conn->query('SELECT * FROM news'); } catch (PDOException $e) { echo 'Error: '.$e->getMessage(); exit; } So far I been using the following to manipulate the data in object form: while ($row = $result->fetchObject()) { echo $row->title; echo $row->text; } My question is, is there a better way? What is the foreach code equivalent to that? I have had no luck keeping the data in object form using foreach statements. Which method is faster, while or foreach? Thank you, any suggestions are highly appreciated. I can get the info from the database fine with this: Code: [Select] $sql2 = "SELECT 1, 2, 3, 4, 5, 6 FROM ratestable ORDER BY id ASC"; $stmt2 = $db->prepare($sql2); $stmt2->execute(); $e2 = $stmt->fetch(); and display it with this: Code: [Select] <?php while($e2 = $stmt2->fetch()) { ?> <input type="text" name="1" maxlength="10" value="<?php echo $e2['1'] ?>" class="rates" /> <input type="text" name="2" maxlength="10" value="<?php echo $e2['2'] ?>" class="rates" /> <input type="text" name="3" maxlength="10" value="<?php echo $e2['3'] ?>" class="rates" /> <input type="text" name="4" maxlength="10" value="<?php echo $e2['4'] ?>" class="rates" /> <input type="text" name="5" maxlength="10" value="<?php echo $e2['5'] ?>" class="rates" /> <input type="text" name="6" maxlength="10" value="<?php echo $e2['6'] ?>" class="rates" /> <?php } ?> How do I update it to the database? Code: [Select] $sql = "UPDATE rates SET title1=?, title2=?, title3=?, title4=?, title5=?, title6=? THE CODE HERE IS WHERE I AM STUCK.... and that's as far as I can get, need to use an array or foreach??????? [/quote] Hi All, First time posting here. I've googled the problem, but can't seem to find a response that's the same. All I want to do is have a list of id numbers and for each id number in the array, submit a MySQL query to retrieve information relating to the id number. When I execute the code below however, I end up with only the last item in the array being printed in the echo statement. Any clues? Thanks, Code: [Select] // get array of ids $ids = getIDs($ids); // loop through input list foreach ($ids as &$id) { getVarDetails($id); } function getVarDetails($local) { $con = mysql_connect('localhost:3306', 'root', '********'); if (!$con) { die('Could not connect: ' . mysql_error()); } // set database as Ensembl mysql_select_db("Ensembl", $con); $result = mysql_query("SELECT * FROM variations WHERE name = '$local' LIMIT 1"); $row = mysql_fetch_array($result) while($row = mysql_fetch_array($result)) { echo $row['name'] . " " . $row['id']; echo "<br />"; } // close connection mysql_close($con); } This topic has been moved to Application Frameworks. http://www.phpfreaks.com/forums/index.php?topic=355772.0 I'm not sure if the title explains it very well, but here is an image of it. I am trying to figure out how to access the option selected (equal, not equal, etc.), along with its input box, for each column (the checkboxes on the right) selected. The purpose of this is to allow a non-programming administrator user to create a custom query to access information. I can find out which checkboxes are selected no problem, but finding the accompanying drop-down option and input box is difficult for me. How do I do this? If you have a suggestion on how to change it, I'm very open. Thank you in advance! Code: [Select] if(!isset($_POST['submit'])) { echo " <form method='post'> <table> <tr> <td valign='top'>SELECT</td> <td valign='top'> <table>"; // LIST ALL COLUMNS IN A TABLE $result = mysql_query("SELECT * FROM table_name") or die(mysql_error()); $rowcount=mysql_num_rows($result); $y=mysql_num_fields($result); for ($x=0; $x<$y; $x++) { echo " <tr> <td> <input type='checkbox' name='fields[]' value='".mysql_field_name($result, $x)."'>".mysql_field_name($result, $x)." </td> </tr>"; } echo " </table> </td> <td valign='top'>FROM allocations</td> <td valign='top'>WHERE</td> <td> <table>"; // LIST ALL COLUMNS IN A TABLE $result = mysql_query("SELECT * FROM table_name") or die(mysql_error()); $rowcount=mysql_num_rows($result); $y=mysql_num_fields($result); for ($x=0; $x<$y; $x++) { echo " <tr> <td> <input type='checkbox' name='column[]' value='".mysql_field_name($result, $x)."'>".mysql_field_name($result, $x)." </td> <td> <select name='operator[]'> <option value='='>equals</option> <option value='<>'>does not equal</option> <option value='>'>greater than</option> <option value='<'>less than</option> <option value='>='>greater than or equal to</option> <option value='<='>less than or equal to</option> <option value='LIKE'>is like</option> </select> </td> <td><input type='text' name='criteria[]'></td> </tr>"; } echo " </table> </td> <td valign='top'><input type='submit' value='Process Query' name='submit' class='submit'></td> </tr> </table> </form>"; } else { echo "<b>Fields to edit:</b> "; foreach ($_POST['fields'] as $fields) { echo $fields,", "; } echo "<br><br>"; echo "<b>Columns to query:</b> "; foreach ($_POST['column'] as $columns) { echo $columns," HERE IS THE SPOT, "; } } hello everyone, I am about to start coding my pages to display results from a database but before i do i want to know information about the following : Is it best to upload images to a database?and display them accordingly? or is it best to use images from a directory? What is most commonly used and or more reliable? Another topic i have trouble finding information on is actually positioning the output from you mysql database, is this practice done with tables?fields and rows? What is this method called? And is there more then one way to go about controlling result layout on your page? Sorry about the 1001 questions , but i am unable to find a clear answer on the topic ..especially question two . Thanks in advance. At the moment I am creating a search function for my website. The approach I have in mind is a pseudo-PHP database. To give an example: A HTML form will submit the results to a PHP file. HTML FORM - Colour: Black PHP RESULT PAGE - if ($_POST['color'] == 'Black') {readfile("./products/black/*.html");} HTML FORM - Price: <$50 PHP RESULT PAGE - if ($_POST['Price'] == '<$50') {readfile("./products/less50/*.html");} The problem here is if there is an item that is black and costs less than $50, then its going to be listed twice. There is probably some code I can write to ommit the listing of duplicate entries, but it is probably going to be messy, so I am wondering if its better to use a centralized MySQL database, rather than a pseudo-PHP database? I've never used MySQL and don't know much about it and this is my first real attempt at using PHP. after cloasing connection of database i still got the values form database. Code: [Select] <?php session_start(); /* * To change this template, choose Tools | Templates * and open the template in the editor. */ require_once '../database/db_connecting.php'; $dbname="sahansevena";//set database name $con= setConnections();//make connections use implemented methode in db_connectiong.php mysql_select_db($dbname, $con); //update the time and date of the admin table $update_time="update admin set last_logged_date =CURDATE(), last_log_time=CURTIME() where username='$uname'limit 3,4"; //my admin table contain 5 colums they are id, username,password, last_logged_date, last_log_time $link= mysql_query($update_time); // mysql_select_db($dbname, $link); //$con=mysql_connect('localhost', 'root','ijts'); $result="select * from admin where username='a'"; $result=mysql_query($result); mysql_close($con); //here i just check after closing data baseconnection whether i do get reselts but i do, why? echo "after the cnnection was closed"; if(!$result){ echo "cont fetch data"; }else{ $row= mysql_fetch_array($result); echo "id".$row[0]."usrname".$row[1]."passwped".$row[2]."date".$row[3]."time".$row[4]; } // echo "<html>"; //echo "<table border='1' cellspacing='1' cellpadding='2' align='center'>"; // echo "<thead>"; // echo"<tr>"; // echo "<th>"; // echo ID; // echo"</th>"; // echo" <th>";echo Username; echo"</th>"; // echo"<th>";echo Password; echo"</th>"; // echo"<th>";echo Last_logged_date; echo "</th>"; // echo "<th>";echo Last_logged_time; echo "</th>"; // echo" </tr>"; // echo" </thead>"; // echo" <tbody>"; //while($row= mysql_fetch_array($result,MYSQL_BOTH)){ // echo "<tr>"; // echo "<td>"; // echo $row[0]; // echo "</td>"; // echo "<td>"; // echo $row[1]; // echo "</td>"; // echo "<td>"; // echo $row[2]; // echo "</td>"; // echo "<td>"; // echo $row[3]; // echo "</td>"; // echo "<td>"; // echo $row[4]; // echo "</td>"; // echo "</tr>"; // } // echo" </tbody>"; // echo "</table>"; // echo "</html>"; session_destroy(); session_commit(); echo "session and database are closed but i still get values from doatabase session is destroyed".$_SESSION['admin']; ?> session is destroyed but database connection is not closed. thanks I have a very tricky php problem here. At least for me. On a page that is editing a job entry. I have a database generated group of checkboxes some of which have been checked when the job was entered. So the script has to do two things generate the complete list of checkboxes and check the ones that have already been selected in the job request. The below script is what I've done trying to figure it out. Description of what I'm trying to do: I'm generating the the boxes with a call to my database for the list of checkboxes. Then I'm building an array with all the possible items that could be checked to compare against. Ok here is where I think my first problem is. I need to make a second query to another table "worklog" in the same database. This is where the list of checked items are stored. What is the best way to do this? A second query seems wrong "or at least not efficient" Code: [Select] <div class="text">Job Type (<a class="editlist" href="javascript:editlist('listedit.php?edit=typelist',700,400);">edit list</a>):</div> <div class="field"> <?PHP $connection=mysql_connect ("localhost", "user", "password") or die ("I cannot connect to the database."); $db=mysql_select_db ("database", $connection) or die (mysql_error()); $query = "SELECT type FROM typelist ORDER BY type ASC"; $sql_result = mysql_query($query, $connection) or die (mysql_error()); $i=1; echo '<table valign="top"><tr><td>'; while ($row = mysql_fetch_array($sql_result)) { $type = $row["type"]; if ($i > 1 && $i % 26 == 0) echo '</td><td>'; else if ($i) echo ''; ++$i; $aTypes = array ("Spec Ad Campaign", "100 x 100 Logo", "100 x 35 (Featured Developer/Broker)", "100 x 40 (Featured Lender)", "120 x 180 (Home Page Auto)", "120 x 45 (Half Tile - Jobs)", "120x60 (Section Sponsor)", "125 x 40 (Profile Page logo)", "135 x 31 (Autos)", "135 x 60 (Autos)", "135 x 60 (Logotile - Jobs)", "150 x 40 (Auto Logo)", "160 x 240 (monster tile)", "160 x 400 (Skyscraper)", "160 x 600 (Tower)", "170x30 (Section Sponsor)", "240 x 180 JPG/GIF Auto Video", "240 x 180 size FLV video ", "300 x 250 (Story Ad)", "300 x 600 (Halfpage)", "468 x 60 (Banner Jpg/Gif)", "728 x 90 (Leaderboard)", "95 x 75 (Sweeps Logo) ", "Agent Profile Page", "Contest", "Creative Change", "Employer Profile", "Holiday Shop", "Home of the Week", "HP Half Banner (234 x 60 Static)", "Mobile Ads (320x53 *300x50*216x36*168x28)", "Newsletter", "Peelback", "Pencil Ad", "Print", "Real Deals", "Resize ad campaign", "Roll-Over Skyscraper", "Site Sponsor logo (170x30)", "Splash Page Production", "Travel Deals Update", "Video Ad", "Web Development", "Web Maintenance", "Web Quote (or Design)"); $dbTypes = explode(',',$row['type']); foreach ($aTypes as $type){ if(in_array($aType,$dbTypes)){ echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type' CHECKED><span style='color:#000;'>$type</span></input><br/>"; }else { echo "<input style='font-size:10px;' name='type[]' type='checkbox' value='$type'><span style='color:#000;'>$type</span></input><br/>"; } } } echo '</td></tr></table>'; ?> </div> I have a standard form that displays users current data from a mysql database once logged in(code obtained from the internet). Users can then edit their data then submit it to page called editform.php that does the update. All works well except that the page does not display the updated info. Users have to first logout and login again to see the updated info. even refreshing the page does not show the new info. Please tell me where the problem is as i am new to php.
my form page test.php
<?PHP
require_once("./include/membersite_config.php");
if(!$fgmembersite->CheckLogin())
{
$fgmembersite->RedirectToURL("login.php");
exit;
}
?>
<form action="editform.php?id_user=<?= $fgmembersite->UserId() ?>" method="POST">
<input type="hidden" name="id_user" value="<?= $fgmembersite->UserId() ?>"><br>
Name:<br> <input type="text" name="name" size="40" value="<?= $fgmembersite->UserFullName() ?>"><br><br>
Email:<br> <input type="text" name="email" size="40" value="<?= $fgmembersite->UserEmail() ?> "><br><br>
Address:<br> <input type="text" name="address" size="40" value="<?= $fgmembersite->UserAddress() ?> "><br><br>
<button>Submit</button>
my editform.php
<?php
$con = mysqli_connect("localhost","root","user","pass");
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
mysqli_query($con,"UPDATE fgusers3 SET name = '".$_POST['name']."', email= '".$_POST['email']."', address= '".$_POST['address']."' WHERE id_user='".$_POST['id_user']."'");
header("Location: test.php");
?>
I appear to be having a simple simon day! my foreach is only displaying the last result in the table as opposed to each row $num_rows = mysql_num_rows($query); echo "There are $num_rows records.<br>"; while($row = mysql_fetch_row($query)) foreach ($row as $field) { echo '<div class="results">'; echo $field; echo '<br></div>'; } And the source Code: [Select] <link href="../styles/clientbox.css" rel="stylesheet" type="text/css"> <body><br> <h3>My Services</h3> <div class="text"> You currently have the following services with us: </div> There are 2 records.<br><div class="results">test1<br></div> </body> </html> My table has 2 rows which the 1st one has a package of test and the second a package of test1 Can anyone help? Thanks In the code I want to proceed to the next step only after stripslashes and strip_tags are completed. How do I put the code below with if? $stripped = array('name', 'location', 'bio'); foreach ( $_POST as $k => $v ) { if ( in_array($k, $stripped) ) { ${$k} = strip_tags($v); } } foreach ( $_POST as $p => $q ) { if ( in_array($p, $stripped) ) { ${$p} = stripslashes($q); } } Thanks, Ruth. Hi All, I essentially am trying to figure out how to print out all of the zip codes that are associated with a search that I am doing, but not list the same zip code more than once. My current foreach statement is as follows and prints out the zip code for each property that is associated with the search: <ul> <?php foreach($properties as $property) { ?> <li><?=$property->$zipcode?></li> <?php } ?> </ul> How would I modify this to print out 1 occurrence of each zip code that is in the search, even if, for example 20 properties have the same zip code? Thanks in advance! P.S. - Tried using an array to work with it but pretty sure my logic was incorrect when putting it together and couldn't get past the errors. Anyways, thanks in advance for the help! I'm trying to figure out why its saying Warning: Invalid argument supplied for foreach() in /home/content/y/a/n/yankeefaninkc/html/defiant/efedmanager/forms/booking.php on line 17. I thought I did the foreach loop right so far. $eventid = $_GET['id']; $query = "SELECT CONCAT_WS(' ', eventnames.eventname, events.label) AS event, DATE_FORMAT(events.bookingdate, '%M %d, %Y') AS bookingdate, events.nummatches AS nummatches FROM events, eventnames WHERE events.event_id = eventnames.id AND events.id = '" . $eventid . "'"; $result = mysqli_query ( $dbc, $query ); // Run The Query foreach ($nummatches as $matchnum) { echo "Match # $matchnum"; echo "<select class=dropdown name=matchtype id=matchtype title=Match Type>"; echo "<option value=0>- Select -</option>"; $query = 'SELECT id, matchtypename FROM matchtypes'; $result = mysqli_query ( $dbc, $query ); // Run The Query while ( $row = mysqli_fetch_array ( $result, MYSQL_ASSOC ) ) { print "<option value=\"".$row['id']."\">".$row['matchtypename']."</option>\r"; } } Hi can any one help please im still very new but hard trying The problem is that i only get 1 result back from the database i need it to loop any ideas please Code: [Select] $query = mysql_query("SELECT * FROM categorys, sub_categorys WHERE categorys.cat_id = sub_categorys.cat_id"); echo "<ul>"; $row[] = mysql_fetch_array($query); foreach($row as $category){ echo "<li>"; echo $category['category']; foreach($row as $sub_category){ if($category['cat_id'] == $sub_category["cat_id"]){ echo "<ul><li>".$sub_category["sub_category"]."</li></ul>"; } } echo "</li>"; } echo "</ul>"; i have a table that echos 10 entries what i want to do is once it echos 5 records it starts a new table. I dont know how to go about this. Could you guys guide me the right way. Hi i am trying to pass values from a cart to a database using foreach. This is my code: Code: [Select] $columns = array('order_id', 'order_item_id', 'order_product_name', 'order_product_dimensions', 'order_product_options', 'order_product_quantity', 'order_product_price', 'order_product_subtotal', 'order_product_vat', 'order_product_delivery','order_product_total','order_product_sku','order_product_image_name','order_product_image_type','order_product_image_path','order_product_image_size','order_product_artwork_link','order_product_artwork_name','order_product_artwork_type','order_product_artwork_path','order_product_artwork_size','order_firstname','order_lastname','order_companyname','order_email','order_tel','order_billing_address','order_billing_address2','order_billing_town','order_billing_county','order_billing_postcode','order_shipping_address','order_shipping_address2','order_shipping_town','order_shipping_county','order_shipping_postcode','order_status','order_username'); $data = array_fill_keys($columns, 'NULL'); foreach($data as $key => $value) { $data[$key] = empty($_POST[$key]) ? 'NULL' : "'".mysql_escape_string($_POST[$key])."'"; $query = "INSERT INTO orders (id, order_created_at, ".implode(", ",$columns).") VALUES (null, NOW(), ".implode(",",$data).')'; } Problem is posting 2 or more products into the database. How do i split the values so it enters each product into the database? today i programmed my own weight calculator. one thing i am having a problem is with trying to make it so if a weight is over a certain number, then the shipping price will increase by X. shipping prices are charged in 20000 gram blocks, so if I have something that is 22000 grams then that will be two blocks. what i need to do is something like (this is a big guess) Quote $row3 = 50; $weighttotal = 22000; if ($weighttotal > 20000) { foreach (20000) { $row3 += $row3;} } } } can it be done on php?? |
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