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PHP - Mysql To Array
firstly i am quite rusty at php/sql so sorry in advance.
I have a php script to produce graphs with the GD library, alls working well, but I want to modify it so the data comes from my database. so I connect to the database in the usuall way but the script uses an array $values = array("23","32","35","57","12","3","36","54","32","15","43","24","30"); so I need mysql to out put in the same format This is what I have so far, it prints out the data with some formating, thank you for any help you can offer Gary Code: [Select] // Make a MySQL Connection $query = "SELECT Question, COUNT(Answer) FROM question1 GROUP BY Question"; $result = mysql_query($query) or die(mysql_error()); // Print out result while($row = mysql_fetch_array($result)){ echo "There are ". $row['COUNT(Answer)'] ." ". $row['Question'] ." items."; } Similar TutorialsHow can i save array from inputs, witch is $igraci and it have 5 values, to mysql base, i tryed but in mysql it shows Array. This is form: <form action="dodaj_klan.php" method="post" > <p> <label>Naziv klana:</label> <input name="naziv" type="text" size="20%" /> <label>Web Sajt:</label> <input name="website" type="text" size="20%" /> <label>E-Mail:</label> <input name="email" type="text" size="20%" /> <br /><br /> <label>Igraci klana(5):</label> <input name="lider" type="radio" value="1" /> <input name="igraci[]" type="text" size="20%" /><br /> <input name="lider" type="radio" value="2" /> <input name="igraci[]" type="text" size="20%" /><br /> <input name="lider" type="radio" value="3" /> <input name="igraci[]" type="text" size="20%" /><br /> <input name="lider" type="radio" value="4" /> <input name="igraci[]" type="text" size="20%" /><br /> <input name="lider" type="radio" value="5" /> <input name="igraci[]" type="text" size="20%" /> <label><i>(obelezi lidera klana)</i></label> <br /><br /> <input class="button" type="submit" name="submit" value="Dodaj" /> </p> </form> Now i wonna save this igraci[] array in mysql, i tried like this but it doesn't work: $igraci = $_POST['igraci']; $query = "INSERT INTO klanovi (naziv, website, email, igraci) VALUES ('{$naziv}', '{$website}', '{$email}', '{$igraci}')"; $result = mysql_query($query, $connection); How can i do this? Thanks.. So I'm querying my database to add the results (mapID's) into a PHP array. The MySQL query I used in the below code would usually return 10 values (only 10 mapID's in the database) Code: [Select] while($data = mysql_fetch_array(mysql_query("SELECT mapID FROM maps"))){ $sqlsearchdata[] = $data['mapID']; } Instead the page takes ages to load then gives this error: Quote Fatal error: Allowed memory size of 8388608 bytes exhausted (tried to allocate 16 bytes) It says the error begins on the first line of the above code. I'm assuming this is not the right way to add the value from the MySQL array into a normal PHP array. Can anyone help me? Hello everyone, Sorry if this has been answered but if it has I can't find it anywhere. So, from the begining then. Lets say I had a member table and in it I wanted to store what their top 3 interests are. Their$ row has all the usual things to identify them userID and password etc.. and I had a further 3 columns which were labled top3_1 top3_2 & top3_3 to put each of their interests in from a post form. If instead I wanted to store this data as a PHP Array instead (using 1 column instead of 3) is there a way to store it as readable data when you open the PHPmyadmin? At the moment all it says is array and when I call it back to the browser (say on a page where they could review and update their interests) it displays 'a' as top3_01 'r' as top3_02 and 'r' as top3_03 (in each putting what would be 'array' as it appears in the table if there were 5 results. Does anyone know what I mean? For example - If we had a form which collected the top 3 interests to put in a table called users, Code: [Select] <form action="back_to_same_page_for_processing.php" method="post" enctype="multipart/form-data"> <input name="top3_01" type="text" value="enter interest number 1 here" /> <input name="top3_02" type="text" value="enter interest number 2 here" /> <input name="top3_03" type="text" value="enter interest number 3 here" /> <input type="submit" name="update_button" value=" Save and Update! " /> </form> // If my quick code example for this form is not correct dont worry its not the point im getting at :) And they put 'bowling' in top3_01, 'running' in top3_02 and 'diving' in top3_03 and we catch that on the same page with some PHP at the top --> Code: [Select] if (isset($_POST)['update_button']) { $top3_01 = $_POST['top3_01']; // i.e, 'bowling' changing POST vars to local vars $top3_02 = $_POST['top3_02']; // i.e, 'running' $top3_03 = $_POST['top3_03']; // i.e, 'diving' With me so far? If I had a table which had 3 columns (1 for each interest) I could put something like - Code: [Select] include('connect_msql.php'); mysql_query("Select * FROM users WHERE id='$id' AND blah blah blah"); mysql_query("UPDATE users SET top3_01='$top3_01', top3_02='$top3_02', top3_03='$top3_03' WHERE id='$id'"); And hopefully if ive got it right, it will put them each in their own little column. Easy enough huh? But heres the thing, I want to put all these into an array to be stored in the 1 column (say called 'top3') and whats more have them clearly readable in PHPmyadmin and editable from there yet still be able to be called back an rendered on page when requested. Continuing the example then, assuming ive changed the table for the 'top3' column instead of individual colums, I could put something like this - Code: [Select] if (isset($_POST)['update_button']) { $top3_01 = $_POST['top3_01']; // i.e, 'bowling' changing POST vars to local vars $top3_02 = $_POST['top3_02']; // i.e, 'running' $top3_03 = $_POST['top3_03']; // i.e, 'diving' $top3_array = array($top3_01,$top3_02,$top3_03); include('connect_msql.php'); mysql_query("UPDATE members SET top3='$top3_array' WHERE id='$id' AND blah blah blah"); But it will appear in the column as 'Array' and when its called for using a query it will render the literal string. a r r in each field instead. Now I know you can use the 'serialize()' & 'unserialize()' funtcions but it makes the entry in the database practically unreadable. Is there a way to make it readable and editable without having to create a content management system? If so please let me know and I'll be your friend forever, lol, ok maybe not but I'd really appreciate the help anyways. The other thing is, If you can do this or something like it, how am I to add entries to that array to go back into the data base? I hope ive explained myself enough here, but if not say so and I'll have another go. Thanks very much people, L-PLate (P.s if I sort this out on my own ill post it all here) My current code lists between 1-6 results from a MySQL array. How do I extract the data from the list and put it into its own variable (eg: $result1, $result2 etc.) ? Code: [Select] $ran_x = rand(1,6); $ran_x = rand(1,6); $x_query = "SELECT * FROM `x` ORDER BY RAND( ) LIMIT $ran_x"; $get_x = mysql_query($appearance_query); echo 'x(s): '; echo '<br />'; while($row = mysql_fetch_array($get_x)) { echo $row['Appearance']; } Any help would be great, Thanks, otester My language table structure looks like : id | english mail_accepted | Your e-mail is accepted. thank_you | Thank you for your visit. What I want to do is, I want to take all these data to an array like : "lang" array And I can use these in the page like Code: [Select] <?php echo "goodbye ".$lang['thank_you']; ?> I know how to take data, lets say Code: [Select] <?php $query = "select * from language.. "; mysql_query($query); ?>But then ?, which function should I use to take the data to my "$lang" array ? Hi Guys, I have been racking my brain all of today trying to get this to work, im pretty new to it all! What im trying to do is use the following code whcih works fine and returns all the correct information in an Array, and insert this into a MySQL DB. Can anyone point me in the right direction? foreach($html->find('div.rsshelper') as $article) { $item['title'] = $article->find('div.prodlistboxbg', 0)->plaintext; $item['link'] = $article->find('.prodimgurl', 0)->href; $item['image_url'] = $article->find('.image-thumb', 0)->src; $item['old_price'] = $article->find('span.text-pricestrike', 0)->plaintext; $item['current_price'] = $article->find('span.text-pricespecial', 0)->plaintext; $item['saving'] = $article->find('span.price-percentage', 0)->plaintext; $articles[] = $item; } print_r($articles); This is the code that generates the Array with the given data. I know how to connect to the DB but how do I get each row into each row of the DB? Thanks Rory Alright, this has been puzzling me for a while now and I think it's time I seek help before my brain turns to mush. I've been modifying a PHP text based game called Legend of the Green Dragon to add more functionality, etc. One of the things I want to add is the ability to add levels higher then 15. The original array looks like this: $exparray = array(1=>100,2=>400,3=>1002,4=>1912,5=>3140,6=>4707, 7=>6641,8=>8985, 9=>11795,10=>15143,11=>19121,12=>23840, 13=>29437,14=>36071,15=>43930, ); Now, I could just hardcode each level, but noo, I want to be able to do it from a module type thing from the admin panel (which I've already gotten that working gratefully) so I've tried writing a few different scripts to take the information from the MySQL database and turn it into an array like above but I've gotten completely lost. This is what I currently have: $expsql = "SELECT * FROM " . db_prefix("myolevels"); $numrows = mysql_num_rows($expsql); for ($i = 1; $i <= $numrows; $i++) { $expresult = mysql_query($expsql); while ($rows = mysql_fetch_assoc($expresult)) { $exparray = array($i=>$rows['myoexpreq']); } } Although when I call the array, it always returns null for the myoexpreq causing your master to look for you all the time (forcing level up when it shouldn't be). Is there any way to create an array like the first with SQL? Thought this might help as well: PHP Warning: "Variable passed to each() is not an array or object" in /home/ileetcom/public_html/logd/lib/experience.php at 35. Call Stack: 2: each(NULL) called from /home/ileetcom/public_html/logd/lib/experience.php on line 35 3: exp_for_next_level(16, "0") called from /home/ileetcom/public_html/logd/village.php on line 131 I'm trying to insert array of posts categories into MySQL. I'm getting category IDs from drop down list.. ERR: Notice: Notice: Array to string conversion in C:\laragon\*********\includes\functions.php on line 477
Array
Array
(
[0] => Array
(
[0] => 4
[1] => 5
)
)
PHP
try {
$sql = "INSERT INTO posts_category_ids (post_id, category_id) VALUES";
$insertQuery = array();
$insertData = array();
foreach ($filter_category as $row) {
$insertQuery[] = '(?, ?)';
$insertData[] = $id;
$insertData[] = $row;
}
if (!empty($insertQuery)) {
$stmt = $this->conn->prepare($sql);
$stmt->execute($insertData); //477
}
}catch (Exception $e){
echo $e->getMessage();
}
}
I have an array of data from a mysql result. The problem I have is that I want to find out what the next row is going to be in advanced. At the moment I have Code: [Select] while($row = mysql_fetch_assoc($rs) { echo 'Order ID: ' . $row['order_id']; } This just echo's out the order_id of the current row. However I want to find out if the order_id on the next row is going to be the same or different. If it's different I will want to echo out something else. Code: [Select] while($row = mysql_fetch_assoc($rs) { echo 'Order ID: ' . $row['order_id']; //If order_id + 1 is different then echo echo 'Total: ' . $row['total'] } Does anyone know a way round this problem? Thanks Hi, I'm wondering if theres a shortcut to a potential problem I have. I'm currently running a query on my website to pull all the fields from a table in my database, for the data to be used on various parts of the page. Usually I would do something as follows Code: [Select] $result = mysql_query("SELECT * FROM table WHERE page='1'"); while ($row=mysql_fetch_array($result)) { $title = $row["title"]; $data = $row["data"]; } And so on and so forth. I would then call the appropriate data by echoing $data for example. However, my table contains a lot more rows than i've mentioned (Around 25 or so). Rather than assigning each to a variable and having a large portion of variable assignments at the top of the page, is there any clever way of putting all of these values inside of an array. So for example, I could call $array_data["title"] or $array_data["data"]?? So it keeps the same key, but puts it inside of an array that I don't have to loop through each time? Hope that makes sense! Thanks, Edd I created a shell script in OS X which forces my Macbook to connect to a MySQL database to see if my laptop was stolen. I can mark the Macbook as stolen in the database, which would provide me with a series of fuctions; one of these functions allows you to take pictures through the iSight camera and upload to an FTP server. The script and database work well, but now I'm trying to make it more user friendly by making it web-based. Here is an example of the code I'm having trouble with: $picture = mysql_query("SELECT picture FROM laptops"); $picrow = mysql_fetch_array($picture); if ( $picrow[0] == 0){ echo "<td>" . "<center><a href='start.php?cmd=start&id=$row[UID]'><img src='images/stopped.gif'></center></a>" . "</td>"; } else { echo "<td>" . "<center><a href='stop.php?cmd=stop&id=$row[UID]'><img src='images/started.gif'></center></a>" . "</td>"; } In the database there is a "1" for active and a "0" for inactive. What it's looking to do is ask to see if it's active or inactive. If it's active, it will have a green button, and inactive with have a red button. Eventually I'd like to be able to activate and deactivate functions by pressing the button, but I need to complete step 1 first. Currently this code does work (except for the links of course); however, I need to specify the number in the array ($picrow[0], $picrow[1], etc.) How do I have this populate automatically from 0 to infinity? I made a field in the database called UID (starting at 1) that automatically increments. I'm not sure if this would help or not. Thanks in advance. Hello guys, Got a question here.. If I have a database table called features with the following columns with example f_id | f_name | f_status 1 | deck | 1 2 | Fireplace | 1 3 | Alarm | 1 I have another table another table called ads that has a column called features and it has the following 1,2,3 (for example) I want to be able to query the db and display all the f_name. Here is what I have so far but it's not working and need some help Code: [Select] // Query the ads table $result3 = mysql_query("SELECT * FROM ads WHERE ad_id='$id2'") or die(mysql_error()); $row3 = mysql_fetch_array( $result3 ); $features = $row3['features']; echo'This is the features ' . $features .''; $feature2 = explode(",", $features); print_r($feature2); echo "test" . $feature2[0]; // piece1 echo $feature2[1]; // piece2 $result = mysql_query("SELECT * FROM features WHERE f_id=$feature2[1]") or die("Sql error : " . mysql_error()); while($row = mysql_fetch_assoc ($result)){ $f_name=$row["f_name"]; echo $f_name; } Hello Can somepne help me with this issue I have a query who is returning some records ID disciplina moduloUfcd idcpDisciplinas anoTurma
58, Comunicação Visual, 8599, 49, 11 60, Comunicação Visual, 134, 49, 10 When i trying to put this into an array with just one line (merge all with number 49 (idcpDisciplinas) and (anoTurma) 11) i can't get the desired output
My sql query is this one
select c.idPlan, cD.disciplina, cM.moduloUfcd, p.Nome, t.Turma, a.Ano, c2.curso,
c.validCron, c.cronograma, c.dataLimite, c.idProfessor,cM.horas,cM.tempos,
c.pdfCronograma, c2.curso, cM.discModulo, cM.idcpDisciplinas, cM.ano as anoTurma
from cpDiscProfessores c
inner join cpDisciplinas cD on c.idcpDisciplinas = cD.idCpDisciplinas
inner join cpModulos cM on c.idCpModulos = cM.idCpModulos
inner join professores p on c.idProfessor = p.idProfessor
left join turmas t on c.idTurma = t.idTurma
inner join anosescolares a on c.idAnoEscolar = a.idAnoEscolar
left join cursos c2 on c.idCursos = c2.idCursos
where a.Estado = 1;
And in php what i have is this
$result = $stmt->fetchAll(PDO::FETCH_ASSOC);
$final = array();
$json = array();
foreach ($result as $row) {
$moduloUfcd= $row['moduloUfcd'];
if (!isset($moduloUfcd[$moduloUfcd])) {
$horas= $row['horas'];
$final[$moduloUfcd]['idPlan'] = $row['idPlan'];
$final[$moduloUfcd]['disciplina'] = $row['disciplina'];
$final[$moduloUfcd]['Nome'] = $row['Nome'];
$final[$moduloUfcd]['Turma'] = $row['Turma'];
$final[$moduloUfcd]['anoTurma'] = $row['anoTurma'];
$final[$moduloUfcd]['curso'] = $row['curso'];
$final[$moduloUfcd]['validCron'] = $row['validCron'];
$final[$moduloUfcd]['cronograma'] = $row['cronograma'];
$final[$moduloUfcd]['dataLimite'] = $row['dataLimite'];
$final[$moduloUfcd]['idProfessor'] = $row['idProfessor'];
$final[$moduloUfcd]['pdfCronograma'] = $row['pdfCronograma'];
$final[$moduloUfcd]['discModulo'] = $row['discModulo'];
$final[$moduloUfcd]['idcpDisciplinas'] = $row['idcpDisciplinas'];
$final[$moduloUfcd]['moduloUfcd'] = array();
}
$final[$moduloUfcd]['moduloUfcd'][] = $row['moduloUfcd'];
$final[$moduloUfcd]['horas'][] = $row['horas'];
$final[$moduloUfcd]['tempos'][] = $row['tempos'];
}
foreach ($final as $moduloUfcd => $cron) {
$json[] = $cron;
}
echo json_encode($json);
What it gives is :
[
{
idPlan: "60",
disciplina: "Comunicação Visual",
Nome: "XXXXXXXXX",
Turma: "11ºO",
anoTurma: "11",
curso: "Técnico de Audiovisuais",
validCron: "-1",
cronograma: "0",
dataLimite: "2020-10-30",
idProfessor: "168",
pdfCronograma: "/XXXXX/pdf/profissionais/cronogramas/Cronograma -Técnico de Audiovisuais-8599.pdf",
discModulo: "8599",
idcpDisciplinas: "49",
moduloUfcd: [
"8599"
],
horas: [
"25"
],
tempos: [
" 34 "
]
}
]
Hi, I have an array that I've set up as a JSON feed API. The array pulls data from a MySQL query, one of the fields in the query ('description') contains text with escape characters. I'm just wondering how I change either the query to remove the escape characters (ie can you strip escape chars in MYSQL?) - or change the array to strip the escape chars?? Code: [Select] //Run query $result = mysql_query(' SELECT DATE_FORMAT(gl_date,"%d-%m-%Y")as "date", gl_venue as "venue", gl_city as "city", gl_postcode as "postcode", gl_text as "description", concat(DAYOFMONTH(gl_date), MONTHNAME(gl_date),".png") AS "imageName" FROM tg_gig_list where gl_date >= curdate() and gl_publish = 1 order by gl_date '); $array = array(); $array['gigs'] = array(); while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { $array['gigs'][] = $row; } $output = json_encode($array); } i want to create array from mysql what is the script for this...? value like $data = array("1" => .0032, "2" => .0028, "3" => .0021, "4" => .0033, "5" => .0034, "6" => .0031, "7" => .0036, "8" => .0027, "9" => .0024, "10" => .0021, "11" => .0026, "12" => .0024, "13" => .0036, "14" => .0028, "15" => .0025); I hope I'm not being a vampire here sucking the life out of you guys, but I can't find any resources in google that explains how to properly use this function in a mysql array. All it does is add a url for any urls added in plaintext. Code: --- <?php function hyperlink ($string) { $string = preg_replace('#(^|\s)([a-z]+://([^\s\w/]?[\w/])*)#is', '\\1<a href="\\2" target="_blank">\\2</a>', $string); $string = preg_replace('#(^|\s)((www|ftp)\.([^\s\w/]?[\w/])*)#is', '\\1<a href="http://\\2" target="_blank">\\2</a>', $string); $string = preg_replace('#(^|\s)(([a-z0-9._%+-]+)@(([.-]?[a-z0-9])*))#is', '\\1<a href="mailto:\\2">\\2</a>', $string); return $string; } // mysql connection $result = mysql_query("SELECT * FROM top_web_articles ORDER BY ID DESC"); echo "<table border='0'>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td class='newstitle' align='left' valign='top' bgcolor='#D6D6D6'>".$row['Category']."</td></tr>"; echo "<td class='newstitle' align='left' valign='top' bgcolor='#D6D6D6'>". $row['ArticleName'] ."</td></tr>"; echo "<td class='newstitle' align='left' valign='top' bgcolor='#D6D6D6'>"hyperlink(. $row['ArticleDescription'] .);"</td></tr>"; echo "<tr><td class='newstitle' align='left' valign='top' bgcolor='#D6D6D6'><a href='". $row['URL'] . "' target='_blank'><img src='http://www.thenewsguys.ca/images/read entire article.png' alt='' width='130' height='11' border='0' /></a></td>"; echo "</tr>"; echo "<td align='left' valign='top'> </td>"; } echo "</table>"; mysql_close($con); ?> --- Error: "Parse error: syntax error, unexpected T_STRING, expecting ',' or ';' in /home/doofyd5/public_html/trevor/playground/displayarticles.php on line 146" (Clearly an error where I'm abusing the heck out of the bolded line. Any ideas on how to properly code this? Hello. I'm new to php and need a little help please. I try to build a class that return an Array of all the table in my db. and be able to get all the data by there col name (like username, password....) but all i got is the last record from the db. here is the class: <?php class db{ function getall($table, $search){ $sql = "SELECT ". $search ." FROM " . $table; $result = mysql_query($sql) or die('Error, insert query failed' . mysql_error()); while($row=mysql_fetch_assoc($result)){ $data = $row; } return $data; } and this is in the html page: $data = $db->getall("users", "*"); foreach($data as $row){ echo $row['fullname']; echo "<BR>"; } can anybody help me with that? I am in the process of writing code for a dynamic form. I have the part of the form written for information to be entered. I am now trying to write code for the modification of the information in the table. I know that I want to get the table data my doing a MySql Select and putting that information into an array. I am not very familiar with arrays and have spent some time looking for a solution online with no solution. When I print_r the results I get what I want but when I do anything else I get Array. What I am actually looking to do is make it so that the table data is in another variable this is the section of code that I am working with. Code: [Select] $mod_form = "<form name='insert_table' id='insert_table' action='ad_add.php' method='post' enctype='multipart/form-data'> <fieldset> <legend>table information</legend> <input name='table' TYPE='hidden' VALUE='" . $table . "' />"; $mod_sql = "SELECT * FROM $table WHERE id = '$id'"; $mod_sql_result = mysql_query($mod_sql); $allowed_ext = array('jpg', 'jpeg', 'png', 'gif'); $input_array = array(); while ($row = mysql_fetch_array($mod_sql_result, MYSQL_NUM)) { $input_array[] = $row; } foreach ($input_array as $input => $data) { print_r($data); } while ($i < mysql_num_fields($mod_sql_result)) { $header = str_replace("_", " ", (mysql_field_name($mod_sql_result, $i))); $name = mysql_field_name($mod_sql_result, $i); (mysql_field_name($mod_sql_result, $i) == 'image' ? $mod_form .= "<p>" . $header . ": <input name='" . $name . "' id='" . $name . "' type='file' /></p>" : $mod_form .= "<p>" . $header . ": <input name='" . $name . "' id='" . $name . "' type='text' value='" . $input . "'/></p>"); $i++; } $mod_form .= " <p> <input type='submit' name='submit_mod' id='submit_mod' /> </p> </fieldset> </form> "; echo $mod_form; There is a "PHP ajax cascading dropdown using MySql" at codestips.com/php-ajax-cascading-dropdown-using-mysql/ I want to use this technique but with a XML or array file instead of mysql database, but my knowledge about mysql is very low. How I can modify this code to catch the categories and products from an array, instead of mysql database? Code: [Select] $connect=mysql_connect($server, $db_user, $db_pass) or die ("Mysql connecting error"); echo '<table align="center"><tr><td><center><form method="post" action="">Category:<select name="category" onChange="CategoryGrab('."'".'ajaxcalling.php?idCat='."'".'+this.value);">'; $result = mysql_db_query($database, "SELECT * FROM Categories"); $nr=0; while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { $nr++; echo "<option value=".'"'.$row['ID'].'" >'.$row['Name']."</option>"; } echo '</select>'."\n"; echo '<div id="details">Details:<select name="details" width="100" >'; $result = mysql_db_query($database, "SELECT * FROM CategoriesDetails WHERE CategoryID=1"); while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo "<option value=".$row['ID'].">".$row['Name']."</option>"; } echo '</select></div>'; echo '</form></td></tr></table>'; mysql_close($connect); ajaxcalling.php is Code: [Select] include("config.php"); $ID=$_REQUEST['idCat']; $connect=mysql_connect($server, $db_user, $db_pass); echo 'Details:<select name="details" width="100">'; $result = mysql_db_query($database, "SELECT * FROM CategoriesDetails WHERE CategoryID=".$ID); while($row = mysql_fetch_array($result, MYSQL_ASSOC)) { echo "<option value=".$row['ID'].">".$row['Name']."</option>"; } echo '</select>'; mysql_close($connect); Hello, I have been spending over 6 hours trying to square this away and I have looked at some tutorials from PHPFreaks... Maybe I'm missing something. Basically I rebuilding the form from bottom up and I thought some check boxes would be easier for some part of the form instead of drop-down menu. Basically this form is submitting query to MySQL: article_add.php (Basically has the form) article_insert.php (Basically insert the data to MySQL) Here is one part of the form where the check box is located: Here is article_add.php <form action="article_insert.php" method="post"> <-- OTHER INPUT BOXES --> <div class="inputfieldtb"><label for="news_cat">News Categories:</label> <? echo "<table>"; echo "<tr>"; $counter = 1; while ($nc = mysql_fetch_array($news_cats)) { $ncat_id = $nc['ncat_id']; $ncat_name = $nc['ncat_label']; echo "<td><div><input type=\"checkbox\" name=\"category[]\" value=\"$ncat_id\" ''/>$ncat_name</div></td>\n"; if (($counter % 4) == 0) { echo "</tr><tr>"; } $counter ++; } echo "</tr></table>"; echo "<br />"; ?> </div> <-- MORE FORMS --> </form> Here is article_insert.php // CONNECT TO DATABASE include("../mods/connect.php"); <--- OTHER DATA INSERT IN MYSQL --> $news_cat = mysql_real_escape_string(implode(',', $_POST['category'])); // <-- Checkbox Data <--- MORE DATA INSERT IN MYSQL --> $query = "INSERT INTO news_articles VALUES ('<--- MORE CONTENT BEING INSERT INTO MYSQL -->,'$news_cat','$news_content',<--- MORE CONTENT BEING INSERT INTO MYSQL -->'')"; mysql_query($query) or die ('Error inserting new data into the data to the database: ' . mysql_error() . ''); The error I'm getting is "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '','','3','0','','','')' at line 1"; So what am I doing wrong? Once I get this resolved, I would like to learn how to display the checkbox / updated the checkbox as well. Thank you in advance for your time and your help with this problem. |
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