|
PHP - Foreach For Iterating Mysql Result, Or Other Alternatives
Now i use a very complex normal for, how i can use a foreach?
Similar TutorialsI am trying to do the following. Except I know that 'return' is not the right method to use, as it stops the script, so what ends up happening is only one row is returned, instead of the three that are there. With return, the data is being passed without being immediately printed, and I end up with the data (but not all of it, because the script stops) in correct place in the page. If I replace return () with echo(), it works fine, in terms of returning the correct data. However, with the way things are setup, if I use echo, the results print at the head of my page. I am using function CreateSideMenu to establish the values for content, and then another function, later on the index.php page, actually creates the page. So what I need is to have something, similar to return (), that passes the information on, but does not immediately print it. Do I make sense? see code below: function CreateSideMenu () { // open CreateSideMenu function include ('/Users/max/Sites/rdbase-llc/hidden/defin/kinnect01.php'); $query = "SELECT content_element_title, content_element_short_text FROM content_main"; $result = mysql_query ($query, $dbc); while ($row = mysql_fetch_array ($result, MYSQL_ASSOC)) { return ("<p>" . $row[content_element_title] . "</h2>\n<p>" . $row[content_element_short_text] . "</p>"); } Thanks ahead of time. This topic has been moved to Ajax Help. http://www.phpfreaks.com/forums/index.php?topic=326703.0 I'm trying to get a foreach loop to one variable that I can insert into MySQL - I've tried a number of things but can't get it working. How do I create one single variable that is created from the foreach? Thanks! Jason Hi, I am new to php, mysql and have made some progress, but recently ran into a wall. I hope someone can help me. I am building a simple links script that has categories and a list of links with website name, url and a brief description. I have two tables-categories and links. I have gotten the back end finished and can add, edit and delete links and categories. The problem is in the front end. I have the script set up so that when I click on a category name a page is dynamically created using ?id=idnumber. and the category name is echoed on the page. But I can't seem to find the solution I need to get the links that have a matching category name to display. I have tried using JOIN, LEFT JOIN, INNER JOIN, and RIGHT JOIN and get the same result, ALL the links in the links table are displayed. I think I need to create a new array and use the foreach loop to accomplish this, but I am not sure. Nor am I sure how to use the foreach even though I have read several tutorials on it. Here is the code I have so far. Any help would be appreciated. Thanks in advance. Code: [Select] <? include "db.php" ?> <?php $id = intval($_GET['id']); // force user input to an int if(!empty($id)) // if the user input is not empty or 0 { $query = "SELECT * FROM categories WHERE id = $id LIMIT 1"; // attempt to select the row for the given id $result = mysql_query($query) or die(mysql_error()); if(mysql_num_rows($result) > 0) // if the categorie was found { $row = mysql_fetch_assoc($result); // fetch the resultset to an array called $row echo $row['categories']; // echo the categories field from the row // etc. } else { die('Error: Bad ID'); // the categories was not found } } else { die('Error: Bad ID'); // the id given was not valid } mysql_close(); ?> <? include "db.php" ?> <?php $sql="SELECT * FROM links,categories ORDER BY links.categories LIMIT 10"; $result=mysql_query($sql); ?> <table width="80%" align="center" border="0" cellspacing="1" cellpadding="0"> <tr> <td> <table width="100%" border="1" cellspacing="0" cellpadding="3"> <tr> <td colspan="3"><strong>List data from mysql </strong> </td> </tr> <tr> <td align="center"><strong>Name</strong></td> <td align="center"><strong>URL</strong></td> <td align="center"><strong>Content</strong></td> </tr> <?php while($rows=mysql_fetch_array($result)){ ?> <tr> <td><? echo $rows['name']; ?></td> <td><a href="<? echo $rows['url']; ?>"><? echo $rows['url']; ?></a></td> <td><? echo $rows['content']; ?></td> </tr> <?php } ?> </table> </td> </tr> </table> <?php mysql_close(); ?> How can I check if a returned mysql value is equal to '' i.e. nothing? I keep getting an error where the page won't load because the returned value is '' so i need to check for it ok this may not be able to be done .... but here goes nothing ... i have a roster that displays member names as a link to thier profile using a form and onclick submit() now what i didnt realize is that the form name will have to increment for each record returned from the query. i am by no means a guru with php so i need some help adding the foreach statement and setting the form name to increment each time what i need: I need this to automatically go +1 for each record returned Code: [Select] <form name="view3" action="./viewinfo.php" method="post"> <a href="#" onclick="document['view3'].submit()"> and i need the correct way to use the foreach statement ... i have read the manual and it just confuzzeled me . here is what i have so far. <?php include './clan_new/config.php'; include './clan_new/access.php'; $db = mysql_connect ($hostname, $username, $password) or die ('Failed to connect to database: ' . mysql_error()); mysql_select_db($database); $query = "SELECT * FROM $member_table WHERE $member_table.rank = 6 ORDER BY $member_table.name ASC"; $result = mysql_query($query) or die ('Failed to query ' . mysql_error()); while ($row = mysql_fetch_assoc($result)) { $steamid = $row['authid']; $name = $row['name']; $email = $row['email']; $fid = $row['fid']; $avatar = $row['avatar']; echo "<tr>"; if ($avatar==""){ echo "<td> </td>"; } else{ echo "<td><img height='35' width='35' src='./clan_new/avatars/$avatar'></td>"; } ?> <form name="view3" action="./viewinfo.php" method="post"> <input type="hidden" id="authid" name="authid" value="<?php echo "$steamid" ; ?>" /> </form> <td align="center"><a href="#" onclick="document['view3'].submit()"> <?php echo "$name" ; ?></a></td> <?php echo "<td align=\"center\"><a class=\"style2\" href=\"mailto:$email\" class=\"style2\">$email</a></td>"; if ($fid==""){ echo ""; } else{ echo "<td align=\"center\">+<a class=\"style2\" href=\"steam://friends/add/$fid\">Friend</a></td>"; } echo "</tr>"; } mysql_free_result($result); mysql_close($db); ?> Hi, I don't know anything about php and Mysql but I found a tutorial for a shopping cart and everything is working. The only thing is they don't have the email part in the totorial so I'm kind of stuck with my file. Anyone know how to email the resul to me via email with this code? <? include("includes/db.php"); include("includes/functions.php"); if($_REQUEST['command']=='update'){ $name=$_REQUEST['name']; $email=$_REQUEST['email']; $address=$_REQUEST['address']; $phone=$_REQUEST['phone']; $result=mysql_query("insert into customers values('','$name','$email','$address','$phone')"); $customerid=mysql_insert_id(); $date=date('Y-m-d'); $result=mysql_query("insert into orders values('','$date','$customerid')"); $orderid=mysql_insert_id(); $max=count($_SESSION['cart']); for($i=0;$i<$max;$i++){ $pid=$_SESSION['cart'][$i]['productid']; $q=$_SESSION['cart'][$i]['qty']; $price=get_price($pid); mysql_query("insert into order_detail values ($orderid,$pid,$q,$price)"); } die('Thank You! your order has been placed!'); } ?> <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>Billing Info</title> <script language="javascript"> function validate(){ var f=document.form1; if(f.name.value==''){ alert('Your name is required'); f.name.focus(); return false; } f.command.value='update'; f.submit(); } </script> </head> <body> <form name="form1" onsubmit="return validate()"> <input type="hidden" name="command" /> <div align="center"> <h1 align="center">Billing information</h1> <table border="0" cellpadding="2px"> <tr><td>Total:</td><td><?=get_order_total()?></td></tr> <tr><td>Name :</td><td><input type="text" name="name" /></td></tr> <tr><td>Address :</td><td><input type="text" name="address" /></td></tr> <tr><td>Email :</td><td><input type="text" name="email" /></td></tr> <tr><td>Phone :</td><td><input type="text" name="phone" /></td></tr> <tr><td> </td><td><input type="submit" value="Place Order" /></td></tr> </table> </div> </form> </body> </html> Im not sure where to post this but since it includes php il post it here instead of in the mysql forum. ok so, i have a table and i get the values using while($row = mysql_fetch_array($result)){ and then echo them in rows. that works fine but i need to add a class to the last row of my table. I would need somehow to fetch the last row of the array and make it echo something different. Any help is appreciated Thank you hello I'm using this code: Code: [Select] $query="SELECT * FROM `second_content` WHERE CHANGED =0 limit 0,1";to query first row data when I want see it and echo it I recived : Code: [Select] Resource id #2can anyone help me ? thank you Hello, i created this script for a client and have ran into an annoying error with the results displaying on a new line for each result instead of side by side, any help is welcome Cheers Code: [Select] <?php $subcat = mysql_real_escape_string(strip_tags(htmlspecialchars(protect($_GET['subcat'])))); $cat = mysql_real_escape_string(strip_tags(htmlspecialchars(protect($_GET['cat'])))); $sql = @mysql_query("SELECT * FROM cakes WHERE category =\"$cat\" AND sub_cat=\"$subcat\" ORDER BY id DESC"); while ($row = mysql_fetch_array($sql)) { $reference = $row['reference']; $image = $row['image']; echo ("<p><img src='./images/cakes/$image' height='289px' width='177px' alt='IMAGE OF CAKE'></img><br />"); echo ("<b>Reference:</b>$reference"."</p>"); } if (!$reference) { echo 'There are no cakes in this category yet.'; } ?> I know the errors only going to be something small i'm missing, but i've been coding all day I have a checkbox when it is checked i garb its value ---meaning the id i take that id and i would like to use it in my where mysql_query statement then I would like to insert the array from the other foreach one after another with a prov meaning provider here is my code print "<select name='prov[]'>"; print '<option value=""></option>'; mysql_connect("localhost", "root", "root") or die(mysql_error()); mysql_select_db("db1") or die(mysql_error()); $query2 = "SELECT * FROM admin"; $result2 = mysql_query($query2) or die(mysql_error()); while($row2 = mysql_fetch_array($result2)) { if ($row2['priv']==prov){print '<option value="'.$row2['user'].'">'.$row2['user'].'</option>';} } print '</select>'; print '</div>'; <?php if (!empty($_POST[submitdeletetic])) { foreach ($_POST["delete"] as $key => $id) foreach ($_POST["prov"] as $prov) { mysql_query("UPDATE tickets SET prov='$prov' WHERE id='".(int)$id."'") or die(mysql_error()); } } ?> Hi. The idea: I want to pull a client record and simply place all the row columns into editable form field text boxes. Once the user reviews and edits a submit button will update the record. What I have: Code: [Select] $Query01 = "SELECT * FROM `CustomerSignups` WHERE `Id` = '$_GET[Id]'"; $Result01 = mysql_query($Query01) or die("Error 01: " . mysql_error()); while ($get_info = mysql_fetch_row($Result01)) { foreach ($get_info as $field) echo "<p>$field</p>"; It displays as expected, a line for each field. How can I get this to pull the field names also and then I can use those as text box input fields...? Hello Masters I need to iterate through multiple $_POST[''] variables that have been submitted by a form, from the user, checking to see if they have been set or left unfilled. Having a tough time finding a tutorial online... I am a php noob... So I though that I would assign short variable names to the $_POST variables and put them into and array(with names for keys) , then use a for loop and isset() to echo any variable's key in which its value has not been set to the user. I have found out that I can not use variables as array values.... What is the best way to do this? Thanks to you in advance sirs. Hi. First timer here. I'm trying to iterate through 5 arrays in a foreach loop. Yep, I know you'll say just use a counter that increments. Sadly, that gives me horrendous results! So, I am using the next & current methods of arrays. Thing is, it gives me bad results ( items that weren't even chosen on the next page ), or it leaves the first couple of items out. I did this : Code: [Select] <?php $ID = array(); $NAME = array(); $PRICE = array(); $QUANT = array(); $checkCount = 0; foreach (array_keys($_POST) as $key) { $$key = $_POST[$key]; $iPos = strpos($key, "ID"); $nPos = strpos($key, "NAME"); $pPos = strpos($key, "PRICE"); $qPos = strpos($key, "QUANT"); $cPos = strpos($key, "CHECK"); if ($iPos !== false) { $ID[] = $$key; } else { //echo "The string ID was not found in the string "; } if ($nPos !== false) { $NAME[] = $$key; } else { //echo "The string NAME was not found in the string "; } if ($pPos !== false) { $PRICE[] = $$key; } else { //echo "The string PRICE was not found in the string "; } if ($qPos !== false) { $QUANT[] = $$key; } else { //echo "The string QUANT was not found in the string "; } if ($cPos !== false) { $checkCount++; if ($checkCount == 1) { echo "Product ID <input type = 'text' name = 'ID" . current($ID) . "' value = '" . current($ID) . "'></br>"; echo "Product Name <input type = 'text' name = 'NAME" . current($NAME) . "' value = '" . current($NAME) . "'></br>"; echo "Product Price <input type = 'text' name = 'PRICE" . current($PRICE) . "' value = '" . current($PRICE) . "'></br>"; echo "Product Quantity <input type = 'text' name = 'QUANT" . current($QUANT) . "' value = '" . current($QUANT) . "'></br>"; } if ($checkCount > 1) { echo "Product ID <input type = 'text' name = 'ID" . next($ID) . "' value = '" . next($ID) . "'></br>"; echo "Product Name <input type = 'text' name = 'NAME" . next($NAME) . "' value = '" . next($NAME) . "'></br>"; echo "Product Price <input type = 'text' name = 'PRICE" . next($PRICE) . "' value = '" . next($PRICE) . "'></br>"; echo "Product Quantity <input type = 'text' name = 'QUANT" . next($QUANT) . "' value = '" . next($QUANT) . "'></br>"; } } else { //echo "The string CHECK was not found in the string "; } } ?> You will see I did all the processing under the CHECK if statement, the reason for this is, it is easier to identify the items ordered because they have been checked on the previous page. Hey guys, Currently Im using: $row = mysql_fetch_array($result) or die(mysql_error()); echo $row['user_family']. " - ". $row['user_registered']; $row['user_family'] = $fam; $_SESSION['family'] = $fam; to take data from a mysql table & set it as SESSION family. However, I cant seem to get this to set. The information IS being taken from mysql because its being echo'd earlier up in the code, but its just not passing to the session. Any ideas? Hi, I'm trying to make a mysql output to a link so the name will be a link so when you hit this link you will get the full information of this mysql input. Can someone point me in the correct direction? Here is how i get the output from my mysql database into my table. Code: [Select] <td>"; echo $row['name']; echo "</td> im getting an error: Warning: mysql_fetch_array() [function.mysql-fetch-array]: The result type should be either MYSQL_NUM, MYSQL_ASSOC or MYSQL_BOTH. in C:\www\library\mysql.class.php on line 353 $result_type = "MYSQL_ASSOC"; $row = mysql_fetch_array($result, $result_type) but if i use (below) it works fine...is the constant() function the right way to what im doing? $result_type = "MYSQL_ASSOC"; $row = mysql_fetch_array($result, constant($result_type)) thank you Hi, I am new to php but not to programming. I created a script on a windows platform which connects to the mysql database and returns the results of a table. A very basic script which I wrote to simply test my connection worked. The script works fine on my windows machine but not on my new mac. On the mac it simply does not display any records at all. I know that the database connection has been established because there is no error but I can not see why the result set is not being displayed on screen, as I said it worked fine on my windows machine. The Mac has mysql (with data) and apache running for php. Please could someone help as I have no idea what to do now? Script below: Code: [Select] $dbhost = 'localhost'; $dbuser = 'root'; $dbpass = 'root'; $conn = mysql_connect($dbhost, $dbuser, $dbpass) or die ('Error connecting to mysql'); $dbname = 'test'; mysql_select_db($dbname); mysql_select_db("test", $conn); $result = mysql_query("SELECT * FROM new_table"); while($row = mysql_fetch_array($result)) { echo $row['test1'] . " " . $row['test2'] . " " . $row['test3']; echo "<br />"; } mysql_close($con); I am seeking to learn more about the noted subject, how to use PHP to allow a user to enter search terms and search a database. I have experimented with this with little results save for errors. Please see code listed below: search.php <? //// filename = search.php <form method="post" action="result.php3"> <select name="metode" size="1"> <option value="row_name1">metode1</option> <option value="row_name2">metode2</option> </select> <input type="text" name="search" size="25"> <input type="submit" value="Begin Searching!!"> </form> ?> results.php //// filename = result.php3 <? $hostname = "mysql7.000webhost.com"; // Usually localhost. $username = "a4542527_root"; // If you have no username, leave this space empty. $password = "*******"; // The same applies here. $usertable = "people"; // This is the table you made. $dbName = "a4542527_test1"; // This is the main database you connect to. MYSQL_CONNECT($hostname, $username, $password) OR DIE("Unable to connect to database"); @mysql_select_db( "$dbName") or die( "Unable to select database"); ?> <? //error message (not found message) $XX = "No Record Found"; $query = mysql_query("SELECT * FROM $usertable WHERE $metode LIKE '%$search%' LIMIT 0, 30 "); while ($row = mysql_fetch_array($query)) { $variable1=$row["row_name1"]; $variable2=$row["row_name2"]; $variable3=$row["row_name3"]; print ("this is for $variable1, and this print the variable2 end so on..."); } //below this is the function for no record!! if (!$variable1) { print ("$XX"); } //end ?> Upon viewing search.php I receive the error message: Parse error: syntax error, unexpected '<' in /home/a4542527/public_html/search.php on line 3 I believe I may be missing something and am a bit lost. Thank-you in in advance for any help or suggestions. ~Matty I am getting the row results from mysql as called, but they appear as a straight line instead of new table row when I echo like: echo "TABLETABLETABLETABLE"; I am looking for: echo "TABLE TABLE TABLE TABLE"; Here is my code: Code: [Select] $brand = $_POST['brand']; $city = stripslashes($_POST['city']); $state = $_POST['state']; $zip_code = stripslashes($_POST['zip_code']); if($city !=""){ $query = "SELECT name, bus_type, street, city, state, zip_code, brand, quantity, price1, price2, date FROM `prices` WHERE city ='".$city."' AND state ='".$state."' AND brand = '".$brand."'"; }else{ $query = "SELECT name, bus_type, street, city, state, zip_code, brand, quantity, price1, price2, date FROM `prices` WHERE zip_code ='".$zip_code."'"; } $result = mysql_query($query); $count=mysql_num_rows($result); if($count==0 && $zip_code !=""){?> <td width="100%" class="style9">Sorry! There are no results for that city, state and brand.</td> <? }else{ while($row = mysql_fetch_array($result, MYSQL_ASSOC)){ ?> <td width="100%" class="style9"><?php echo "<table width=\"100%\" border=\"1\" cellspacing=\"0\" cellpadding=\"0\"> <tr> <td{$typ_image}</td> <td><b><a href=\"\" title=\"{$row['street']} {$row['city']}, {$row['state']}. {$row['zip_code']}\" target=\"_blank\">{$row['name']}</a></b><br>Last Updated:{$row['date']}</td> <td><b>{$row['brand']} - {$row['quantity']}</b><br>Before Tax:\${$row['price1']} After Tax:\${$row['price2']}</td> <td{$pago}</td> </tr> </table><BR>"; ?></td> <?php } } ?></table> <p> </p> </td> |
|||||||