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PHP - Html Form Value Returning Null In Php Script? (basic Question)
Hello,
I'm trying to take the value from an HTML form and insert it into a database on a button click. It inserts a null value into the database. The script is called submitColumnDetails.php. This is where I create the text field that I want to take the information from. This is in a separate file. Code: [Select] echo <<<END <form action = "submitColumnDetails.php" method = "POST"> <input type = "text" name = "columnField"/> </form> END; This is the submitColumnDetails.php file Code: [Select] <?php //Submit Column Data //-----------------------------------------------------// //Connect to localhost server $connector = mysql_connect("localhost", "root", "root"); if(!$connector){ //If user can't connect to database die('Could not connect: ' . mysql_error()); //Throw an error } //-----------------------------------------------------// mysql_select_db("colin_db", $connector); $newValue = $_POST["columnField"]; //Data from column field. THIS IS WHAT RETURNS NULL $newColumnQuery = "INSERT INTO `colin_db`.`allColumnFields` (`DATA`) VALUES ('$newValue')"; mysql_query($newColumnQuery); //Query to add form to main table $newColumnIntoMainTableQuery = "ALTER TABLE colin_table ADD ('$newValue' varchar(50))"; mysql_query($newColumnIntoMainTableQuery); //Query to add column to main table mysql_close($connector); //Close database connection echo "<script type = 'text/javascript'> window.location = 'index.php'</script>"; //Go back to original page ?> Even when I print out the $newValue, it does not print anything. What am I doing incorrectly? Similar TutorialsOkay so I created a theme for my website that will also utilize a custom CMS script I developed and its not cooperating with me nicely. Here's my head of my header file for my template: Code: [Select] <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Strict//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-strict.dtd"> <html xmlns="http://www.w3.org/1999/xhtml" <?php language_attributes(); ?>> <head> <title><?php bloginfo('name') ?>: <?php bloginfo('description') ?></title> <link rel="stylesheet" type="text/css" media="all" href="<?php bloginfo( 'stylesheet_url' ); ?>" /> <meta http-equiv="content-type" content='<?php bloginfo("html_type"); ?> charset=<?php bloginfo('charset'); ?>' /> <?php if(is_singular()){ wp_enqueue_script('comment-reply');} ?> <?php wp_head(); ?> <?php require ('efedmanager/inc/dbconfig.php'); ?> </head> Now on my index page that calls the header file I DO NOT get a error saying that the file could not be located. So that's good news. Here's my dbconfig.php file. Code: [Select] <?php /** * @author Jeff Davidson * @copyright 2010 */ // This file contains the database access information and establishes a connection to MySQL and selects the database // Set the database access information as contstants DEFINE ('DB_USER', '?'); DEFINE ('DB_PASSWORD', '?'); DEFINE ('DB_HOST', '?'); DEFINE ('DB_NAME', '?'); // Make the database connection $dbc = mysqli_connect (DB_HOST, DB_USER, DB_PASSWORD, DB_NAME); if (!$dbc) { die('Connect Error (' . mysqli_connect_errno() . ') ' . mysqli_connect_error()); } ?> Now when I go to one of my custom pages like this one it brings up the message: Warning: mysqli_query() expects parameter 1 to be mysqli, null given in /home/xtremer/public_html/wp-content/plugins/exec-php/includes/runtime.php(42) : eval()'d code on line 38 Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, null given in /home/xtremer/public_html/wp-content/plugins/exec-php/includes/runtime.php(42) : eval()'d code on line 39 Code: [Select] <div id="champions" class="content"> <h1 class="pageheading">KOW Champions and Contenders</h1> <?php $championsQuery = " SELECT titles.titleName, titles.shortName, champions.champID, champions.con1ID, champions.con2ID, champions.con3ID, biochamp.shortName AS championshortName, biochamp.characterName AS champion, biocon1.shortName AS con1shortName, biocon1.characterName AS con1, biocon2.shortName AS con2shortName, biocon2.characterName AS con2, biocon3.shortName AS con3shortName, biocon3.characterName AS con3 FROM champions LEFT JOIN titles AS titles ON titles.ID = champions.titleID LEFT JOIN characters AS biochamp ON champions.champID = biochamp.ID LEFT JOIN characters AS biocon1 ON champions.con1ID = biocon1.ID LEFT JOIN characters AS biocon2 ON champions.con2ID = biocon2.ID LEFT JOIN characters AS biocon3 ON champions.con3ID = biocon3.ID WHERE titles.statusID = '1' ORDER BY titles.ID"; $championsResult = mysqli_query($dbc, $championsQuery); while ( $row = mysqli_fetch_array ( $championsResult, MYSQLI_ASSOC ) ) { $fieldarray=array('titleName','shortName','championID','championshortName','champion','con1ID','con1','con1shortName','con2ID','con2','con2shortName','con3ID','con3','con3shortName'); foreach ($fieldarray as $fieldlabel) { ${$fieldlabel} = $row[$fieldlabel]; } ?> <div id="title"><span class="large">< ?php echo $titleName ?></span> < ?php if (file_exists('images/championshots/'.$titleshortName.'/'.$championshortame.'.png')) { echo "<img class=champion src=images/championshots/".$shortName."/".$champion.".png alt= />\n"; } else { echo "<img class=champion src=images/championshots/".$shortName."/".$shortName.".png alt= />\n"; } if (strlen ($champion) < 1) { echo "<span class=medium>Vacant"; } else { echo "<span class=medium><a href=/bio?shortName=".$championshortName.">".$champion."</a></span>\n"; echo "<span class=medium>(Since TBD)</span>"; } ?> <span class="contender">Contenders</span> <ul> < ?php if ( strlen ($con1) < 1) { echo "<li><span class=medium>TBD</span>"; } else { echo "<li><a href=/bio?shortName=".$con1shortName.">".$con1."</a></li>\n"; } if (strlen ($con2) < 1) { echo "<li><span class=medium>TBD</span>"; } else { echo "<li><a href=/bio?shortName=".$con2shortName.">".$con2."</a></li>\n"; } if (strlen ($con3) < 1) { echo "<li><span class=medium>TBD</span>"; } else { echo "<li><a href=/bio?shortName=".$con3shortName.">".$con3."</a></li>\n"; } ?> </ul> </div> < ?php } ?> </div> Hello
I am having an issue with encoding an array to JSON, my array seems to be formed correctly but when I try and echo the json_encode version I get "null". I echo the JSON error code and it is "0" indicating no issue..
I have read that non UTF-8 characters can be an issue, but this is simple data, no strange characters... but I added the utf8_encode anyway....
Here is my code:
$dataArray = array(
'title' => 'Test Item',
'icon_url' => 'http://url/api/testimage.jpg',
'item_id' => '12345',
'expiration' => 600
);
utf8_encode($dataArray);
$responsArray = array(
'success' => '1',
'data' => $dataArray
);
utf8_encode($responsArray);
header('Content-type: application/json');
echo print_r($responsArray);
echo json_encode ($responseArray);
echo json_last_error();
Trying to figure out how to better write this chunk of code. I'm wanting to get the list of the roster members and then create an array of options for the view dropdown to display inside the select dropdown and also have it have an option to display "Please Select An Option". However what if what is returned from the getAllRoster function is NULL which is what I have returned if no results are returned from a query. How should I handle that which I just want the empty option displayed. Also I need to think about is do a function to retrieve all the allies for that specific matter and then display that ally as the default ally in the dropdown for each dropdown. Controller: Code: [Select] $rosterList = $this->bios->getAllRoster(); $allies = array(); $allies[''] = 'Please Select An Opion'; foreach ($rosterList AS $ally) { $allies[$ally->id] = $ally->rosterName; } View: Code: [Select] <?php echo form_label( 'Ally 1', 'ally1'); ?> <div> <?php echo form_dropdown( 'ally1', $allies, ''); ?> </div> <?php echo form_label( 'Ally 2', 'ally2'); ?> <div> <?php echo form_dropdown( 'ally2', $allies, ''); ?> </div> <?php echo form_label( 'Ally 3', 'ally3'); ?> <div> <?php echo form_dropdown( 'ally3', $allies, ''); ?> </div> Hi all,
I'm posting JSON to an API call. If I use a var_dump within the API it will return correctly, although if I try to var_dump within the function it will return NULL.
API:
$json = file_get_contents('php://input');
if(isset($_GET['form_id']) && intval($_GET['form_id'])>0)
{
if(isset($json))
{
$records = $questionnaire->logFormAnswers($_POST['answers'],$_GET['form_id']);
Then using the logFormAnswers function...
Function:
public function logFormAnswers($answers = "", $form_id = 0)
{
$answers = json_decode($json,true);
var_dump($answers);
The JSON:
[{"question_id":"3","answer":"dsf"},{"question_id":"3","answer":"sdfsdfdsfsd"}]
Eventually, once the JSON has been decoded, it will be inserted into a database within the function.
Ok, this may be just because I have been programming all day and my mind has gone blank (happens alot), but this is my PHP script: Code: [Select] <?php $query_distinct_item_types = mysql_query("SELECT DISTINCT name FROM item_types"); while($item_types = mysql_fetch_array($query_distinct_item_types)){ $distinct_item_types[] = $item_types['name']; } foreach($distinct_item_types as $item){ $query_item_total = mysql_query("SELECT item_type, SUM(price) WHERE item_type='$item' FROM costs GROUP BY item_type"); while($item_total = mysql_fetch_array($query_total_price)){ $item_totals[] = $item_total['SUM(price)']; } } $item_summery = $item_totals; ?> $item_summery which is = to $item_totals is returning null, any idea's? Hello, I'm writing a PHP program in which textfields are printed (in a loop) into an HTML table. On a button click, another column is added to the table, and another textfield added. The problem is that when a new textfield is printed, it is slightly above the height of the other textfields (see attached picture if unclear). The code for this is below: Code: [Select] <table border = '1' cellspacing = '4' style = "color: white"> <?php include "connectToDatabase.php"; //Initial script call to generate table from database ?> </table> //Code in connectToDatabase.php... $query = mysql_query("SELECT * FROM colin_table"); while($row = mysql_fetch_array($query)){ //$row represents each row in an array for($i = 0; $i < $limit; $i++){ //Print every element in the array if($row[$i] == null) echo "<td align = 'center'><input type = 'text' style = 'text-align: center; vertical-align: middle' align = 'center'>"; else echo "<td align = 'center'>" . $row[$i]; } echo "<td align = 'center'>"; echo "<tr align = 'center'>"; //Then move to the next row } //Loop that prints text boxes for every column in the HTML table $textBoxQuery = mysql_query("SELECT * FROM allColumnFields"); while($row = mysql_fetch_array($textBoxQuery)){ echo <<<END <td align = 'center'> <input type = 'text' style="margin-left:auto; text-align:center; vertical-align: middle; margin-right:auto;" name = "rowField"/> END; //THIS IS THE PART THAT NEEDS WORK. FOR SOME REASON NEW TEXTBOX IS HIGHER THAN THE OTHER BOXES } It's more of an aesthetic problem, but it's nonetheless something I'd like to get fixed. Any ideas? Jake I'm trying to get this to either show "Final" or "Final Overtime" but it just continues to show "Final" even when the XML key value says Overtime...
What am I missing here?
Thank you!
Hello folks, I've got a problem with a search script. It's searching a db containing various properties (as in buildings) and it's not quite returning the correct results. The search form asks that you select a house type (4 options, all of which are tick boxes) and then the last part of the form is select how many bedrooms you'd like (which is a drop-down list featuring numbers 2-5) What should happen is someone selects one or more of the tick boxes and then the number of bedrooms and it returns all the house types that were ticked but only those with the number of bedrooms that were specified in the original search. Currently the results are correct in that they're showing the correct house types, but the number of bedrooms isn't right. e.g. if you chose 2 bedrooms in the search it shows those, as well as 3, 4 and 5 bedrooms in the results. Can't figure it out but it will be something to do with && and OR in my search script. Can anyone suggest a tweak to this that might work? (I'm using the $_GET function because the search form is in a Flash movie) $Result = mysql_query("select * from property_names where TownHouse = '$_GET[TownHouse]' OR Apartment = '$_GET[Apartment]' OR Detached = '$_GET[Detached]' OR SemiDetached = '$_GET[SemiDetached]' && visible= 'Yes' && bedrooms = '$_GET[Bedrooms]' && bedrooms = '$_GET[Bedrooms]'") Hello, freaks. I need a little help with the best way to go about this. I have a field in mysql that contains a comma separated string of values. What I need to do is get those values out of the db and list them into editable text fields so the user can update the values in a html form. So far this is my code, I left out all the in betweens - validation, results, etc - for this post...but its there . SELECT servs FROM services WHERE id = $a_id $services = $row['servs']; $srvs = explode(',', $services); Then for the form: <?php foreach ($srvs as $service) { ?> <input type="text" name="servs[]" value="<?php echo $service ?>" /> <?php } ?> This gives me a text field for each value of servs, great! But the problem is, the user can enter up to 15 servs. Some currently have less than that saved in the db. I want to give them the ability to 1. edit any of the current servs value and 2. add more values if they have less than 15. My question is, how do I echo out 15 editable text fields for servs[] no matter the number of values they currently have? And when I go to implode the values, will blank text fields screw everything up? I can't find anything in the forums that relate to this. I believe it probably will involve $i, but I am not too familiar with counting in php yet, so I could really use some help. Thanks! Oh - and I am not stuck on using text fields, but I couldn't figure how to make this work using a select list or checkboxes. Any suggestions either way would help. Hi guys, sorry for such a newbish question. Any help would be greatly appreciated. HTML FORM: Code: [Select] <form action="form.php" method="post" onsubmit="return validateForm()" name="form"> <b>First Name:*</b> <input type="text" name="first_name" size="50" /> <b>Last Name:*</b> <input type="text" name="last_name" size="50" /> <b>Phone:*</b> <input type="text" name="phone" size="50" /> <b>Email:*</b> <input type="text" name="email" size="50" /> <p><b>What is your favorite color?*</b></p> <p align="left"> <select name="se"> <option value="W">White</option> <option value="G">Green</option> <option value="Y">Yellow</option> </select> <input type="submit" value="Submit"/> </form> FORM.PHP script Code: [Select] <?php $se = $_POST['se']; $seURL = ''; switch ($se) { case 'W': $seURL = "http://url1.com"; break; case 'G': $seURL = "http://url2.com"; break; case 'O': $seURL = "http://url3.com"; break; default: $seURL = ""; } if ($seURL != "") { /* Redirect browser */ /* make sure nothing is output to the page before this statement */ header("Location: " . $seURL); } // get posted data into local variables $EmailFrom = "noreply@domain.com"; $EmailTo = "email@domain.com"; $Subject = "Form"; $first_name = Trim(stripslashes($_POST['first_name'])); $last_name = Trim(stripslashes($_POST['last_name'])); $phone = Trim(stripslashes($_POST['phone'])); $email = Trim(stripslashes($_POST['email'])); // validation $validationOK=true; if (!$validationOK) { print "<meta http-equiv=\"refresh\" content=\"0;URL=error.htm\">"; exit; } // prepare email body text $Body = ""; $Body .= "first_name: "; $Body .= $first_name; $Body .= "\n"; $Body .= "last_name: "; $Body .= $last_name; $Body .= "\n"; $Body .= "phone: "; $Body .= $phone; $Body .= "\n"; $Body .= "email: "; $Body .= $email; $Body .= "\n"; $Body .= "color: "; $Body .= $se; $Body .= "\n"; // send email $success = mail($EmailTo, $Subject, $Body, "From: <$EmailFrom>"); // send email to user if ($se=="W") $EmailFrom = "noreply@domain.com"; $to = $email; $subject = "form email"; $body = "thank you for filling out our form"; if (mail($to, $subject, $body, "From: <$EmailFrom>")) { echo("<p>Message successfully sent!</p>"); } else { echo("<p>Message delivery failed...</p>"); } ?> [code] MOD EDIT: [nobbc][code] . . . [/code][/nobbc] tags added . . . Code: [Select] //policies subreport function opportunity_sub_report($business_contact_id){ global $edit, $styles; $sql="Select * from opportunities WHERE op_business_contact_id='$business_contact_id'"; $rs=myload($sql); echo '<table>'; if (count($rs)>0){ // print a header row echo '<tr> <td>Opportunity ID</td> <td>Category</td> <td>Agent</td> <td>Opportunity Name</td> <td>Priority</td> <td>Assigned To</td> <td>Opportunity Type</td> <td>Opporunity Status</td> <td>Opportunity Submission</td> <td>Due Date</td> <td>Contracting Recieved(Y or N)?</td> <td>New Business</td> <td>Notes</td> <td>Business Contact ID</td> </tr>'; // print all data rows foreach ($rs as $r){ echo '<tr> <td>'.display_field('', $p,$r['op_id']).'</td> <td>'.display_field('',$p,$r['op_category']).'</td> <td>'.display_field('', $p,$r['op_agent_id']).'</td> <td>'.display_field('', $p, $r['op_opportunity_name']).'</td> <td>'.display_field('', $p, $r['op_priority']).'</td> <td>'.display_field('', $p, $r['op_assigned_to']).'</td> <td>'.display_field('', $p, $r['op_opportunity_type']).'</td> <td>'.display_field('', $p, $r['op_status']).'</td> <td>'.display_field('', $p, $r['op_submission_date']).'</td> <td>'.display_field('', $p, $r['op_due_date']).'</td> <td>'.display_field('', $p, $r['op_contracting_received_y_or_n']).'</td> <td>'.display_field('', $p, $r['op_new_business_app_received_y_or_n']).'</td> <td>'.display_field('', $p, $r['op_wholesaler_notes']).'</td> <td>'.display_field('', $p, $r['op_business_contact_id']).'</td> </tr>'; } } echo '</table>'; } I'm having some trouble with some code I'm tasked with maintaining. This code works, but it doesn't return the information in a tab. I'm using tabber.js It's being called here Code: [Select] echo ' <tr> <td class="timegrid" colspan=5> <div class="tabber"> <div class="tabbertab" title="Contracting"></div> <div class="tabbertab" title="Licensing"></div> <div class="tabbertab" title="Cases"></div> <div class="tabbertab" title="Commissions"></div> <div class="tabbertab" title="Opportunities">'.opportunity_sub_report($rs[0][bc_business_contact_id]).'</div> <div class="tabbertab" title="Tasks"></div> <div class="tabbertab" title="Calendar"></div> <div class="tabbertab" title="Attachments">'.agent_detail_attachments($rec_id).'</div> </div> </td> </tr>'; echo '</table>'; I've tried changing it from echo to return, but I'm not sure how to keep the foreach loop intact in such a way. I'm trying to create subreports in a web app, that would act like access. Any help would be much appreciated. Hi, I m doing some work for my self an because of that i been reading a lot arround about PHP, and theres something that i would like to ask a bit of enlightenment. So my question is as the title says about html form's using php to insert data into mysql, i been reading tutorials arround the interwebs and even made afew successful tests, but pretty much all tutorials use 2 files to accomplish this the html file with the form and an insert.php where the actual code is stored so this made me think is this how usually it's done? in over all you will have 1 file for the form, 1 for the insert, 1 for the edit php code and 1 for delete. How do you guys usually do it? PS: one of the tests i did was making 1 single file with all these using an switch. My interest in making this question is solo to learn how other people do it to see if i m in the right way. Thanks in advance. is it okay to name a column "date" in a table ??? just want to know whether this could mess up everything thanks in advance Hello, Im from the Netherlands and i have a little question about php. I have a form in php where administrators can make new users. They can fill in a name, password, age and can select a role. The role is a dropdown menu where they can select athlete, researcher and trainer. My problem is that when they select a role, i want that selected role transfert to a php variable. I want to use the variable in a if statment, when sporter selected a new dropdown appears with options like tennis, hockey.... I have seen a lot on google about this with jquery and stuff, but i cant find a good example. Can u help me? Thanks. Quote from: http://www.killerphp.com/tutorials/object-oriented-php/php-objects-page-1.php getter and setter names should match the property names. This way, when other PHP programmers want to use your objects, they will know that if you have a method/function called 'set_name()', there will be a property/variable called 'name'. Is this an accurate statement? Even if there's like 20+ properties? That would be 40+ methods and that's not including doing anything with those properties but setting/getting. That sounds...I don't know...against the grain. I would like to keep all my classes in a folder placed within the htdocs folder where my php script runs. Let's say the classes folder is com\jimslounge\classes. From my script can I instantiate a class like this $myClass = new com\jimslounge\classes\Jimsclass(); Does it work something like that? I would hate to have to use include to access classes. Hi, can you help me please? I have some code written about 5 yrs ago that is for example like this echo"<div class=\"select\">"; echo"<br /><br />\n"; echo"</div>\n"; echo"</div>\n"; echo "<p><b>$address</b></p>\n"; Does it make any difference at all if I amended it to echo"<div class=\"select\"><br /><br /></div></div><p><b>$address</b></p>"; ie coding errors or increase load time? also when should I use \n"; or "; Please forgive my ignorance but I am a self taught php writer and have managed to teach myself but just need a little help. Cheers Does it matter where you start your session at? Does it have to be before the <!doctype> tag or can it go anywhere in the page? I have a php string variable that is created by php code within an html form ($answer). I need to pass this string variable along with all the html form input data to another php script specified with the form "action" (post method). All the html form input data is coming thru fine but not the variable ($answer). How do I do this? Here is the php code for importing html form data at the script called in the form action: $languages = $_POST['languages']; $answergiven = $_POST['answergiven']; $problemanswer = $_POST['$answer']; 'languages' and 'answergiven' are form inputs and come thru fine. '$answer' does not get passed to the second script. How do I do this? Here is the php code within the first html form <?php // OPEN DATABASE $username="servics3_sample"; $password="sample"; $database="servics3_sample"; mysql_connect(localhost,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); // GENERATE RANDOM PROBLEM NUMBER $probnum = (rand ( 1 , 9 )); echo $probnum; // RETRIEVE ANTI-SPAM PROBLEM $query="SELECT * FROM liasantispam WHERE `problem number` LIKE '%$probnum%' "; $result=mysql_query($query) or die("Error: ". mysql_error(). " with query ". $query); $firstnum=mysql_result($result,0,"first number"); $operator=mysql_result($result,0,"operator"); $secondnum=mysql_result($result,0,"second number"); $answer=mysql_result($result,0,"answer"); echo $firstnum," ",$operator," ",$secondnum," = "; mysql_close(); ?> |
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