PHP - Select Different Srts Of Values
Hi all i have 2 small questions and maybe someone could help me out.
Say I have a database with a table 'users'. This table has fields: email, name and year. What I want to do is show at the front-end all users, by selecting them from the table, and arrange them by year in column 1 and show them as individuals in column 2. But here is the thing; if someone selects for instance year 1970 and also someone from 1970 individually, someone will be selected twice. (edit:As a side note the selecting is aimed and emailing the users, that's why selecting someone twice is not good First question is: does anyone know how to remove doubles (maybe even triples if there is a select all button, maybe turn this around and say select only 1 instance of a user). Second question: is this a wise way to let someone select. Right now all users are fetched, and if the table users grows this could be tricky because of the large number of users. If someone could point me in the right direction of thinking I would be very happy, been thinking about this and scalability for quite a while now Similar Tutorials
Table Issue - Multiple Location Values For User Pushes Values Out Of Row Instead Of Wrapping In Cell
I have a Prepared Statement that runs a SELECT statement and returns 2 records, and I would like to store the Field Value for each Record into an Array. Here is the code I usually use for queries that return just a single value... // ****************** // Populate Form. * // ****************** // Build query. $q2 = "SELECT response FROM bio_answer WHERE member_id=?"; // Prepare statement. $stmt2 = mysqli_prepare($dbc, $q2); // Bind variable to query. mysqli_stmt_bind_param($stmt2, 'i', $memberID); // Execute query. mysqli_stmt_execute($stmt2); // Store results. mysqli_stmt_store_result($stmt2); // Check # of Records Returned. if (mysqli_stmt_num_rows($stmt2)>0){ // Details Found. // Bind result-set to variable. mysqli_stmt_bind_result($stmt2, $response); // Fetch record. mysqli_stmt_fetch($stmt2); // Close prepared statement. mysqli_stmt_close($stmt2); }else{ // Details Not Found. $_SESSION['resultsCode'] = 'DETAILS_NOT_FOUND_2133'; // Close prepared statement. mysqli_stmt_close($stmt2); // Set Error Source. $_SESSION['errorPage'] = $_SERVER['SCRIPT_NAME']; // Redirect to Display Outcome. header("Location: " . BASE_URL . "/members/results.php"); // End script. exit(); }//End of POPULATE FORM Can someone help me out with the syntax so that I get an end result like this... $answerArray[0] = 'I want to be my own boss!!' $answerArray[1] = 'Don't waste your time trying to do your own Taxes!' Thanks, Debbie Hi all, Currently my dob_day is not showing any value, even though there is a value in the database, it seems like it is unable to retrieve the value. No value was shown in the <select> box. Do you guys have any idea? Thanks Code: [Select] <?php $query = "SELECT name, nric, gender, picture, dob_day FROM tutor_profile WHERE tutor_id = '" . $_GET['tutor_id'] . "'"; $data = mysqli_query($dbc, $query) or die(mysqli_error($dbc)); // The user row was found so display the user data if (mysqli_num_rows($data) == 1) { $row = mysqli_fetch_array($data); if ($row != NULL) { $name = $row['name']; $nric = $row['nric']; $gender = $row['gender']; $old_picture = $row['picture']; $dob_day = $row['dob_day']; } else { echo '<p class="error">There was a problem accessing your profile.</p>'; } //HTML CODING <label for="dob" class="label">Date Of Birth</label> <select id="dob_day" name="dob_day" value="<?php if (!empty($dob_day)) echo $dob_day; ?>"> <option> <option>01</option> <option>02</option> <option>03</option> <option>04</option> <option>05</option> <option>06</option> <option>07</option> <option>08</option> <option>09</option> <option>10</option> <option>11</option> <option>12</option> <option>13</option> <option>14</option> <option>15</option> <option>16</option> <option>17</option> <option>18</option> <option>19</option> <option>20</option> <option>21</option> <option>22</option> <option>23</option> <option>24</option> <option>25</option> <option>26</option> <option>27</option> <option>28</option> <option>29</option> <option>30</option> <option>31</option> </select> ?> Hi guys, I am trying to create a script that gets a multiple select form selections and assign them to variables with php via the POST method. This is my code: Code: [Select] <select name="search_commodity[]" multiple="multiple" size="3" style="position:absolute; top:80px; left:414px; " > <option value="all">All</option> <option value="none">None</option> <option value="some">Some</option> </select> Then after the form gets submited then it should go to this php page and the first value of the chosen options has to be assigned to the variable called commo. The problem is that is is not assigning anything. Code: [Select] <?php $commo = $_POST['search_commodity[0]']; ?> Can someone help please? I can't figure it out. hi people can vote 3 times with 3 select menus but i want them not to be able to choose 3x the same select option the options in the select menus come out of the database this is how i let them vote Code: [Select] <?php include("./includes/egl_inc.php"); $secure = new secure(); $secure->secureGlobals(); session_start(); if (isset ($_GET['match'])) { $matchid = (int)$_GET['match']; $_SESSION['matchid'] = $matchid; $matchmaps = mysql_fetch_array(mysql_query("SELECT maps,game FROM ffa_matches WHERE id=$matchid")); $matchmapschecker = $matchmaps[maps]; $matchgamechecker = $matchmaps[game]; $maps=mysql_query("SELECT id,map FROM ffa_maplists where gameid = $matchgamechecker"); while(list($id,$map)=mysql_fetch_row($maps)) { $maplist.="<option value='$id'>$map</option>\n"; } $out[body].="<table width='98%' border='0' cellspacing='0' cellpadding='0' style='border:1px solid black'> <tr style='color:#cccccc; background-color:black;'> <td valign='middle' background='$config[bg2]' align='center' colspan='10'> <br/>Choose Maps<br/> </td> </tr></table><table width='98%' border='0' cellspacing='0' cellpadding='0' style='border:1px solid black'> <tr style='background-color:black;'> <td> <form action='ffainsertmapvote.php' method='post'> <select name='map1'>$maplist</select> <select name='map2'>$maplist</select> <select name='map3'>$maplist</select> <input type='hidden' name='match' value='$matchid' /> <input type='submit' value='Vote' /> </form> </td> </tr></table>"; } include("$config[html]"); ?> do i need jquery to fix this? i dont want the user to be able to go to a next screen without having max 2 the same options selected thanks in advance ! Hi guys, Im developing a stock system and so far new stock and can be added with the quantity, however im trying to update the stock by using a select box and selecting the type of stock and inputting a new qty. the form and everything is set up and i can select files from the database, however i dont know how to update files from a select box The code i got so far is Code: [Select] <!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>stock manager</title> </head> <body> <?php $stockqty =&$_POST['stock_qty1']; ?> <center> <table> <td> <table> <td> <form action='stockview.php' method='POST'> Please Enter a Stock Name and Stock Value <table> <tr> <td> Stock Name: </td> <td> <input name="stock_name" type="text" /><BR /> </td> </tr> <tr> <td> Stock Qty: </td> <td> <input name="stock_qty" type="text" /> </td> </tr> </table> <input name="submit" type="submit" value="Add New Stock Items" /> </form> </td> </table> </td> <td> <table> <td> <form action='stockmanager.php' method='POST' enctype="multipart/form-data"> Please Select from the list the item you wish to update <table> <tr> <td> Stock Name: </td> <td> <?php $connect = mysql_connect("localhost","root", "") or die ("Couldn't Connect!"); mysql_select_db("stock", $connect) or die("Couldn't find db"); // select database $query=("SELECT id, stockname FROM stocks"); $result = mysql_query ($query); echo "<select name=stock value=''>Edit Stock QTY</option>"; while($nt=mysql_fetch_array($result)) { //Array or records stored in $nt echo "<option value=$nt[id]>$nt[stockname]</option>"; /* Option values are added by looping through the array */ } $queryreg = mysql_query("SELECT * FROM stocks WHERE stockqty='$stockqty'"); $numrows = mysql_num_rows($queryreg); $update = mysql_query("UPDATE stocks SET stockqty='$stockqty' WHERE stockname = '$stockname'"); echo("Its updated"); ?> </td> </tr> <tr> <td> Stock Qty: </td> <td> <input name="stock_qty1" type="text" /> </td> </tr> </table> <input name="submit" type="submit" value="Update stock items" /> </form> </td> </table> </td> </table> </center> </body> </html> Any help would be greatly appreciated. Thanks Lance hi all please help... i would like to get datas between certain intervals. please have a look @ db_table.jpg, it's the db structure. the red highlighted part is the value i need. condition is select values when both device one & device two are off. db_table_report.jpg is the resulting report format. It contains start time & end time and the duration.. please help, i know it's a bit complicated, but it's urgent. I have set up a form page with a select box of colleges to select. I want the "options" in the select box to be values taken from a field called "name" in a table called "colleges" and they should be ordered alphabetically. I also want the default selected option to be "none." I have attached a picture to describe what i want. Please be detailed with the code. I am fairly new to php and mysql. Thank you. Hi, In my mysql database i have a text input option, in the registration form and edit my details form i have a multiple select dropdown list, which user selects options to populate the text input box, which ultimately populates the text field in the mysql database. All works perfectly. The dropdownlist consists of 3 parts <optgroups> first is current selection (what is the usesr current selection)works fine, The second <optgroup> is existing words, what words we(the site) have given as options, and the third <optgroup> is the words that others have used. This is where im having a small problem. Because its a text field when i call the data from the database, it calls the entire text box as a single option in my select list.. I want to break the words in the text field (at the comma) and have them listed each one as an option in the select list. Example what i need: Words in text box:(my input allows the "comma") word1, word2, word3, word4, word5, word6, How i want them called/displayed: <option value=\"word1\">word1</option> <option value=\"word2\">word2</option> <option value=\"word3\">word3</option> <option value=\"word4\">word4</option> <option value=\"word5\">word5</option> <option value=\"word6\">word6</option> here's my code: $query = "SELECT allwords FROM #__functions_experience WHERE profile_id = '".(int)$profileId."' LIMIT 1"; $original_functionsexperience =doSelectSql($query,1); $query = "SELECT allwords FROM #__functions_experience WHERE profile_id = '".(int)$profileId."' LIMIT 1"; $functionsexperiencelist=doSelectSql($query); $funcexpList ="<select multiple=\"multiple\" onchange=\"setFunctionsexperience(this.options)\">"; foreach ($functionsexperiencelist as $functionsexperienceal) { $selected=""; if ($functionsexperienceals->allwords == $original_functionsexperience) $selected=' selected="selected"'; $allwords=$functionsexperienceal->allwords; $funcexpList .= "<optgroup label=\"Current selection\"> <option value=\"".$allwords."\" ".$selected." >".$allwords."</option> </optgroup> <optgroup label=\"Existing Words\"> <option value=\"existing1,\">existing1</option> <option value=\"existing2,\">existing2</option> <option value=\"existing3,\">existing3</option> <option value=\"existing4,\">existing4</option> <option value=\"existing5,\">existing5</option> <option value=\"existing6,\">existing6</option> </optgroup> <optgroup label=\"Others added\"> //heres problem <option value=\"".$allwordsgeneral."\">".$allwordsgeneral."</option> </optgroup>"; } $funcexpList.="</select>"; $output['FUNCEXPLIST']=$funcexpList; The result im getting for optgroup others added: word1, word2, word3, word4, word5, how can i get it like this: <option value=\"word1\">word1</option> <option value=\"word2\">word2</option> <option value=\"word3\">word3</option> <option value=\"word4\">word4</option> <option value=\"word5\">word5</option> <option value=\"word6\">word6</option> Dear All Members here is my table data.. (4 Columns/1row in mysql table)
id order_no order_date miles How to split(miles) single column into (state, miles) two columns and output like following 5 columns /4rows in mysql using php code.
(5 Columns in mysql table) id order_no order_date state miles 310 001 02-15-2020 MI 108.53 310 001 02-15-2020 Oh 194.57 310 001 02-15-2020 PA 182.22
310 001 02-15-2020 WA 238.57 ------------------my php code -----------
<?php
if(isset($_POST["add"]))
$miles = explode("\r\n", $_POST["miles"]);
$query = $dbh->prepare($sql);
$lastInsertId = $dbh->lastInsertId(); if($query->execute()) {
$sql = "update tis_invoice set flag='1' where order_no=:order_no"; $query->execute();
} ----------------- my form code ------------------
<?php -- Can any one help how to correct my code..present nothing inserted on table
Thank You Edited February 8, 2020 by karthicbabuHi, My company has 240+ locations and as such some users (general managers) cover multiple sites. When I run a query to pull user information, when the user has multiple sites to his or her name, its adds the second / third sites to the next columns, rather than wrapping it inside the same table cell. It also works the opposite way, if a piece of data is missing in the database and is blank, its pull the following columns in. Both cases mess up the table and formatting. I'm extremely new to any kind of programming and maybe this isn't the forum for this question but figured I'd give it a chance since I'm stuck. The HTML/PHP code is below: <table id="datatables-column-search-select-inputs" class="table table-striped" style="width:100%"> <thead> <tr> <th>ID</th> <th>FirstName</th> <th>LastName</th> <th>Username</th> <th>Phone #</th> <th>Location</th> <th>Title</th> <th>Role</th> <th>Actions</th> </tr> </thead> <tbody> <?php //QUERY TO SELECT ALL USERS FROM DATABASE $query = "SELECT * FROM users"; $select_users = mysqli_query($connection,$query);
// SET VARIABLE TO ARRAY FROM QUERY while($row = mysqli_fetch_assoc($select_users)) { $user_id = $row['user_id']; $user_firstname = $row['user_firstname']; $user_lastname = $row['user_lastname']; $username = $row['username']; $user_phone = $row['user_phone']; $user_image = $row['user_image']; $user_title_id = $row['user_title_id']; $user_role_id = $row['user_role_id'];
// POPULATES DATA INTO THE TABLE echo "<tr>"; echo "<td>{$user_id}</td>"; echo "<td>{$user_firstname}</td>"; echo "<td>{$user_lastname}</td>"; echo "<td>{$username}</td>"; echo "<td>{$user_phone}</td>";
//PULL SITE STATUS BASED ON SITE STATUS ID $query = "SELECT * FROM sites WHERE site_manager_id = {$user_id} "; $select_site = mysqli_query($connection, $query); while($row = mysqli_fetch_assoc($select_site)) { $site_name = $row['site_name']; echo "<td>{$site_name}</td>"; } echo "<td>{$user_title_id}</td>"; echo "<td>{$user_role_id}</td>"; echo "<td class='table-action'> <a href='#'><i class='align-middle' data-feather='edit-2'></i></a> <a href='#'><i class='align-middle' data-feather='trash'></i></a> </td>"; //echo "<td><a href='users.php?source=edit_user&p_id={$user_id}'>Edit</a></td>"; echo "</tr>"; } ?>
<tr> <td>ID</td> <td>FirstName</td> <td>LastName</td> <td>Username</td> <td>Phone #</td> <td>Location</td> <td>Title</td> <td>Role</td> <td class="table-action"> <a href="#"><i class="align-middle" data-feather="edit-2"></i></a> <a href="#"><i class="align-middle" data-feather="trash"></i></a> </td> </tr> </tbody> <tfoot> <tr> <th>ID</th> <th>FirstName</th> <th>LastName</th> <th>Username</th> <th>Phone #</th> <th>Location</th> <th>Title</th> <th>Role</th> </tr> </tfoot> </table>
hirealimo.com.au/code1.php this works as i want it: Quote SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1 but apparelty selecting * is not a good thing???? but if I do this: Quote SELECT priceID, price FROM price INNER JOIN vehicle....etc it works but i lose the info from the vehicle table. but how do i make this work: Quote SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc so that i am specifiying which colums from which tables to query?? thanks Hi all, Just curious why this works: Code: [Select] while (($data = fgetcsv($handle, 1000, ",")) !== FALSE){ $import="INSERT into $prodtblname ($csvheaders1) values('$data[0]','$data[1]','$data[2]','$data[3]','$data[4]','$data[5]','$data[6]')"; } And this does not: $headdata_1 = "'$data[0]','$data[1]','$data[2]','$data[3]','$data[4]','$data[5]','$data[6]'"; while (($data = fgetcsv($handle, 1000, ",")) !== FALSE){ $import="INSERT into $prodtblname ($csvheaders1) values($headdata_1)"; }it puts $data[#'s] in the database fields instead of the actual data that '$data[0]','$data[1]'... relates to. I wrote a script to create the values in $headdata_1 based on the number of headers in $csvheaders1 but can't seem to get it working in the sql statement. Thanks Hi all, I'm a first time poster here and I would really appreciate some guidance with my latest php challenge! I've spent the entire day googling and reading and to be honest I think I'm really over my head and need the assistance of someone experienced to advise the best way to go! I have a multi dimensional array that looks like (see below); the array is created by CodeIgniter's database library (the rows returned from a select query) but I think this is a generic PHP question as opposed to having anything to do with CI because it related to working with arrays. I'm wondering how I might go about searching the array below for the key problem_id and a value equal to a variable which I would provide. Then, when it finds an array with a the matching key and variable, it outputs the other values in that part of the array too. For example, using the sample data below. How would you recommend that I search the array for all the arrays that have the key problem_id and the value 3 and then have it output the value of the key problem_update_date and the value of the key problem_update_text. Then keep searching to find the next occurrence? Thanks in advance, as above, I've been searching really hard for the answer and believe i'm over my head! Output of print_r($updates); CI_DB_mysql_result Object ( [conn_id] => Resource id #30 [result_id] => Resource id #35 [result_array] => Array ( ) [result_object] => Array ( ) [current_row] => 0 [num_rows] => 5 [row_data] => ) Output of print_r($updates->result_array()); Array ( [0] => Array ( [problem_update_id] => 1 [problem_id] => 3 [problem_update_date] => 2010-10-01 [problem_update_text] => Some details about a paricular issue [problem_update_active] => 1 ) [1] => Array ( [problem_update_id] => 4 [problem_id] => 3 [problem_update_date] => 2010-10-01 [problem_update_text] => Another update about the problem with an ID of 3 [problem_update_active] => 1 ) [2] => Array ( [problem_update_id] => 5 [problem_id] => 4 [problem_update_date] => 2010-10-12 [problem_update_text] => An update about the problem with an ID of four [problem_update_active] => 1 ) [3] => Array ( [problem_update_id] => 6 [problem_id] => 4 [problem_update_date] => 2010-10-12 [problem_update_text] => An update about the problem with an ID of 6 [problem_update_active] => 1 ) [4] => Array ( [problem_update_id] => 7 [problem_id] => 3 [problem_update_date] => 2010-10-12 [problem_update_text] => Some new update about the problem with the ID of 3 [problem_update_active] => 1 ) ) I have 2 queries that I want to join together to make one row
Dear All, I wish to have 2 drop down boxes, Country Select Box and Locality Select Box. The locality select box will be affected by the value chosen in the country select box. All is working fine except that the locality select box is not being populated. I know that the problem is in the sql statement WHERE country_id='$co' because i am having an error that $co is an undefined variable. All the rest works fine because i have replaced the $co variable directly with a number (say 98) for a particular country id and it worked fine. In what way can i define this variable $co so that it is accepted by my sql statement? Thank you for your help in advance. MySQL Tables indicated below: CREATE TABLE countries( country_id INT(3) UNSIGNED NOT NULL AUTO_INCREMENT, country_name VARCHAR(30) NOT NULL, PRIMARY KEY(country_id), UNIQUE KEY(country_name), INDEX(country_id), INDEX(country_name)) ENGINE=MyISAM; CREATE TABLE localities( locality_id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT, country_id INT(3) UNSIGNED NOT NULL, locality_name VARCHAR(50), PRIMARY KEY (locality_id), INDEX (country_id), INDEX (locality_name)) ENGINE=MyISAM; Extract PHP script included below: // connect to database require_once(MYSQL); if(isset($_POST['submitted'])) { // trim the incoming data /* this line runs every element in $_POST through the trim() function, and assigns the returned result to the new $trimmed array */ $trimmed=array_map('trim',$_POST); // clean the data $co=mysqli_real_escape_string($dbc,$trimmed['country']); $lc=mysqli_real_escape_string($dbc,$trimmed['locality']); } ?> <form action="form.php" method="post"> <p>Country <select name="country"> <option>Select Country</option> <?php $q="SELECT country_id, country_name FROM countries"; $r=mysqli_query($dbc,$q) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); while($row=mysqli_fetch_array($r)) { $country_id=$row[0]; $country_name=$row[1]; echo '<option value="' . $country_id . '"'; if(isset($trimmed['country']) && ($trimmed['country']==$country_id)) echo 'selected="selected"'; echo '>' . $country_name . '</option>\n'; } ?> </select> </p> <p>Locality <select name="locality"> <option>Select Locality</option> <?php $ql="SELECT locality_id, country_id, locality_name FROM localities WHERE country_id='$co' ORDER BY locality_name"; $rl=mysqli_query($dbc,$ql) or trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc)); while($row=mysqli_fetch_array($rl)) { $locality_id=$row[0]; $country_id=$row[1]; $locality_name=$row[2]; echo '<option value="' . $locality_id . '"'; if(isset($trimmed['locality']) && ($trimmed['locality']==$locality_id)) echo 'selected="selected"'; echo '>' . $locality_name . '</option>\n'; } // close database connection mysqli_close($dbc); ?> </select> </p> <p><input type="submit" name="submit" value="Submit" /></p> <input type="hidden" name="submitted" value="TRUE" /> </form> I'm trying to run a readout of a db which runs fine as an individual script. When I embed it in PHP inside an html div container, it bails when it encounters the first ">" and simply outputs the PHP characters from there through the "?>". The rest of the html runs fine before and after. For example, this line echo "<select>"; would output "; and any other php script up to the ?> end, after which it renders html fine. If I run the script as a separate php file, it runs as expected. Any help would be appreciated. Thanks. Hello,
i am starting to freak out ... i am trying to make a select but it dont want to work.
So what do i have:
i have 2 Tables
table 1: customer (contains the name) (fieldname: name)
table 2: bookings (contains the name from customer) (fieldname: belegung)
in table 2 there are multiple lines with lets say 10 customer xyz and 3 customer xxx
in table 1 every customer is only once and it includes 5 customer
my output i want to have now is a group by where i have the customer in column 1 and the count(*) in column 2
so in my case
customer xyz - 10
customer xxx - 3
customer 3 - 0
customer 4 - 0
customer 5 - 0
i only manage to group them, so i can see it like
customer xyz - 10
customer xxx - 3
customer 3 - 1
customer 4 - 1
customer 5 - 1
as it goes on the count(*) from table 1
please if someone could help me i would be pleased
I am trying to get an id from a table where clientid is equal to a session var. I know the session is set because if I echo it, it shows up, and I know there is a row in the db that matches because I am looking at it right now. The problem is, the query is not selecting any data. I need to select the clientid from the table so I can set another session var. Please please help my brain is not coping! Thank you! <? session_start(); $uid = $_SESSION['logid']; include('../connectdb.php'); mysql_select_db("maindb", $con); $result = mysql_query("SELECT * FROM projects WHERE clientid = $uid"); while($row = mysql_fetch_array($result)) { echo $row['id']; } ?> I'm down to just echoing the row to problem solve but nothing comes up. Hi, I'm not quite sure how to do this, so i thought i'd ask you guys from some assistance. Basically i am inserting Author's into a table successfully using an array. The reason for this is that i have multiple authors being added and there is no limit as to how many. An example of what i mean can be seen he http://www.prima.cse.salford.ac.uk:8080/~ibrarhussain/test.html You can click on "Add author" to add however many necessary.. Anyway, i have got the insert working, however when i edit i want to be able to see all the authors that have been added but obviously i don't know how many there are.. Typically i would like to see something like this: http://www.prima.cse.salford.ac.uk:8080/~ibrarhussain/edit.jpg So i would click on an edit link and it would pre populate the text boxes. I don't have a problem with doing this, but how can i show the correct amount of input textboxes based on how many authors exist for that specific record? Can someone offer some advice please? The input elements are like so: Quote <input type="text" name="author[]" id="author1"/> <input type="text" name="author[]" id="author2"/> <input type="text" name="author[]" id="author3"/> ... ... ... <input type="text" name="author[]" id="author10"/> Some records may have 1 author some may have 10, so how can i do this? Thanks again.. Hi Guys
I have a table which in it's shortened form has the following columns:
id | postID | title | content | version
The column for postID has a number that can be shared by multiple rows - differentiated by version number.
I want to run a query to select all records that are like a given keyword (i.e. %LIKE%) but where results share the same postID I only want to return the highest version number for that record.
The difficulty is some records may have multiple version numbers that match the like statement and some may have only one. So this variance with the LIKE search is causing me some confusion.
I've tried this in a few ways using a sub-query but for the life of me I cannot work out how to do it.
Any help would be appreciated,
Drongo
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