PHP - Variable Not Changing When It Should - From Null
if ($rowcats->followersrange == "2000-5000") { $lowrange2000 = "2000"; $lowestrange = $lowrange2000; $numberpeople = $rowcats->numberpeople; } if ($rowcats->followersrange == "5000-10000") { $lowrange5000 = "5000"; if ($lowrange5000 < $lowestrange) { $lowestrange = $lowrange5000; $numberpeople = $rowcats->numberpeople;} } if ($rowcats->followersrange == "10000-15000") { $lowrange10000 = "10000"; if ($lowrange10000 < $lowestrange) { $lowestrange = "$lowrange10000"; $numberpeople = $rowcats->numberpeople;} if ($lowestrange == NULL) { echo "yes";} echo "$lowestrange"; } if ($rowcats->followersrange == "15000-20000") { $lowrange15000 = "15000"; if ($lowrange15000 < $lowestrange) { $lowestrange = $lowrange15000;$numberpeople = $rowcats->numberpeople;} } We are trying to find the lowest range from the entries added to a database, so at the end, we can say the range is FROM... TO... They might opt for 2-5k, or 5-10k etc. If they opt for 10-15k, then the lowest range is 10,000. We have a result of 10000 - 15000. So the $lowrange10000 = "10000". Therefore, this should be echoing the $lowestrange from 7th line up in this code, but it remains at NULL, based on: $lowestrange = NULL;, set at the top of the code. Why is that? Similar TutorialsHow can I check a variable to see if it is NULL and if so set to 0000-00-00? I've tried: if ($var IS NULL) THEN $var='0000-00-00'; to no avail. Okay, so, I have a script that allows a forum inside of a room in a game. If you want a forum, you mark Yes from a dropdown menu inside the editing portion of the room. It will then save that room's title into a field called `board` in the database. However, when you mark No from the dropdown, it still leaves it in there. I don't want this. I want it to change to NULL. How do I go about doing this? Here's my current coding: if (!islevel($p,$co)) newlevel($p,$co); setlevel($p,$co,"title",filter(stripslashes($title))); setlevel($p,$co,"board",filter(stripslashes($title))); } Explanations: setlevel just connects to the database and inserts these values. $title is obviously the title of the room $co = coordinates for the room Here is the editing page, where the dropdown menu is: <table width=100% style=text-align:center><tr> <td>BOARD:<br /> <select name=b> <option value=0>No</option> <option value=1 <? if ($b == 1) echo "selected" ?>>Yes</option></select><br />Was <?=$b?> </tr></table> Been scratching my head about this all day. Cannot figure it out. Any help would be appreciated. Code: [Select] <?php $selectClause = 'SELECT * FROM idInfo WHERE'; $whereClause = ''; function addWhere($fxnName, $fxnValue) { if (!$whereClause) { return $fxnName . '=\'' . $fxnValue . '\''; } else { return ' AND ' . $fxnName . '=\'' . $fxnValue . '\''; } } $whereClause.=addWhere("firstName", $firstName); $whereClause.=addWhere("lastName", $lastName); $whereClause.=addWhere("idNum", $idNum); echo 'mysql_query("' . $selectClause . ' ' . $whereClause . '"); <br /><br />'; ?> This is meant to generate a query of the type: Code: [Select] SELECT * FROM idInfo WHERE firstName='firstNameINPUT' AND lastName='lastNameINPUT' AND idNum='idNumINPUT' For some reason, though, the if statement always reads false. No matter how I try to work the syntax Code: [Select] if (!$whereClause) if ($whereClause) if ($whereClause == NULL) if ($whereClause != NULL) if (empty($whereClause) etc. I've tried all of those and more, experimenting with and without quotes. They always return a FALSE for the existence of the variable, so that the string never contains the word "AND" as it should by the end. I've tried these same if statements outside the function and they work fine. Inside the function, it doesn't matter what the value of the variable is, the function runs as though it has no value. Comments? Suggestions? I am facing problem to execute query by assigning NULL value to a variable and then executing query.In MySQL DB four fields Mobile,landline, pincode,dob are set as integer and date(for dob) respectively.The default is set as NULL and NULL option is selected as yes.All these fields are not mandatory.The problem is that when I edit the form my keeping the value as empty in DB these are saved as 0, 0 , 0 & 0000-00-00 inspite of Null. I have tried everything but still the defect persist. Please help me to come out of the problem The code, I have used: <?php //require_once 'includes/config.php'; $dbusername = $_POST['email']; $dbfirstname = $_POST['first_name']; $dblastname = $_POST['last_name']; //$dbmobile_number = $_POST['mobile']; if (isset($_POST['mobile'])) { $dbmobile_number = $_POST['mobile']; } else { $dbmobile_number = "NULL"; } $dblandline_number = $_POST['landline']; $dbdob = $_POST['dob']; if(isset($_POST['is_email'])) { $dbSubscribe_Email_Alert = '1'; } else { $dbSubscribe_Email_Alert = '0'; } if(isset($_POST['is_sms'])) { $dbSubscribe_SMS = 0; } else { $dbSubscribe_SMS = 0; } $dbAddress_firstname = $_POST['shipping_first_name']; $dbAddress_lastname = $_POST['shipping_last_name']; $dbAddress = $_POST['shipping_address']; $dbcity = $_POST['shipping_city']; $dbpincode = $_POST['shipping_pincode']; $dbstate = $_POST['shipping_state']; $dbcountry = $_POST['shipping_country']; echo "Welcome".$dbusername; //if($_POST['btnSave']) //if ($_POST['btnSave']) //{ //echo "Inside query loop"; $connect = mysql_connect("localhost","root","") or die("Couldn't connect!"); mysql_select_db("salebees") or die ("Couldn't find DB"); //$query = mysql_query("SELECT * FROM users WHERE username='$username'"); $query = mysql_query("update users set firstname = '$dbfirstname', lastname = '$dblastname', mobile_number = '$dbmobile_number', landline_number = '$dblandline_number', dob = '$dbdob', Subscribe_Email_Alert = '$dbSubscribe_Email_Alert', Subscribe_SMS = '$dbSubscribe_SMS', Address_firstname = '$dbAddress_firstname', Address_lastname = '$dbAddress_lastname', Address = '$dbAddress', city = '$dbcity', pincode = '$dbpincode', state = '$dbstate', country = '$dbcountry' where username = '$dbusername' "); header("location:my_account.php"); //} //else //{ //die(); //} ?> Hi, I have a weather app, coded which creates weather based on a set zip code. I want to create a form which when the user fills out their local code and clicks the send button that will change the zip code variable. Currently it is set up like this $zip = '90210'; //change to your zipcode need help with getting started with this. Thanks! Hello everyone, I'm having a difficult time figuring out what my problem is here. I'm trying to submit a couple of strings (to the user they are messages to be sent to other users) and I am having trouble with the string being cut off at quotation marks. Here is the code that I'm using: Code: [Select] <?php if (isset($_POST['submit'])) { $staffmessage = $_POST['staffmessage']; $studentmessage = $_POST['studentmessage']; echo "$staffmessage<br /> <br />$studentmessage"; } ?> <!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=iso-8859-1" /> </head> <body> <?php $message1="This is a test message. It does not get cut off here when the \"quotation marks\" are added."; $message2="This is a test message. It DOES get cut off here when the \"quotation marks\" are added."; ?> <form name="form" method="post" action="test.php"> <textarea class="textarea" name="staffmessage" rows="12"> <?php echo $message1; ?> </textarea></p> <input name="studentmessage" type="hidden" value="<?php echo $message2; ?>" /> <input class="submit2" type="submit" name="submit" value="Send Message" /> </form> </body> </html> The result that I'm getting is that when $staffmessage is echoed, I get the full message. When $studentmessage is echoed, the message gets cut off at the first quotation mark. The only thing that is different when creating these two variables is that $staffmessage is submitted using a "textarea" field in the form, and $studentmessage is submitted using a "hidden" field in the form. Other than that, they are handled the exact same way. Can anyone please help me figure out how to remedy this so that $studentmessage is not cut off at the quotation mark? Thanks in advance for your help! I am trying to get a simple thing working. I have a domain with google that I want to change the point to IP when I access a PHP page. I have a DB table with settings that I am using for the username, password, etc. The issue is that the command to change the IP always sends the same IP. No matter what I try. I have tried assigning a fixed entry for the variable $localip, I have set that variable to 0 ahead of the query to get the info from the table. Nothing I do will change the variable in the command. I can manually put the command in a browser and it will perform the change. Can anyone comment on what I am missing here?
<?php ini_set("display_errors", 1); ini_set("track_errors", 1); ini_set("html_errors", 1); error_reporting(E_ALL); require('/var/www/html/cqadmin/utils/connect.php'); $sql = "SELECT * FROM `failover` LIMIT 0, 30"; $result = mysqli_query($link,$sql) or die(mysql_error()); $row = mysqli_fetch_array($result); $localip = ($row['localip']); $dnsusername = ($row['dnsusername']); $dnspassword = ($row['dnspassword']); $domain = ($row['failoverurl']); $location = ($row['location']); $dealer = ($row['dealer']); $output = shell_exec("curl -s https://$dnsusername:$dnspassword@domains.google.com/nic/update?hostname=$domain&myip=$localip"); mysqli_close($link); ?> I get no errors on the page and if I echo $output I get a nochng message back from Google
Thanks for looking!! Usually it would be something like:
$("#lblpasswordmessage").html("Password is too short");But how do you do it using a variable? For example say the variable is called 'pwm'; $("#pwm").html("Password is too short"); $("pwm").html("Password is too short"); $(pwm).html("Password is too short");Thanks I am trying to update the database with isset to set the value of the variable to either null or the value entered by the user. Currently, if no value has been entered null is being set as a string causing the date field to populate with zeros. How do I code it so that it populates or sets the value as null and not as string null?
Here's my code: (yes I know sql injections it's still in development )
<?php Hello All, I tried searching here and google, but couldn't figure it out on my own... still Bellow is my code, I need some help figuring out how to tell PHP that "If picture is NULL, don't try to display it." Basically I have a dynamic news system set up. Some stories have pictures, some do not. I need PHP to not display the code for showing the picture if the picture value = NULL. Any help would be appreciated! <?php include ("../../includes/connections/newsconnection.php"); $id=$_GET['id']; //declare the SQL statement that will query the database $query = " SELECT * FROM news WHERE id='$id' "; //execute the SQL query and return records $result = mssql_query($query); //display the results //id, school, date, link, title, story, picture while($row = mssql_fetch_array($result)) { echo "<span class='bodyb'>"; echo $row["title"]; echo "</span>"; echo "<br /><i>"; echo date('l, F j, Y', strtotime($row['date'])); echo "</i>"; echo "<br /><br />"; //heres where I need to stop if picture is null echo "<img src='../../images/news/"; echo $row["picture"]; echo "' align='left' class='imgspacer-left'>"; echo fixQuotes($row["story"]); } //close the connection mssql_close($dbhandle); ?> Can someone help me to get this code working... Code: [Select] public function getBowlContents(){ if is_null($this->contents) { return "The bowl is empty!"; } else { return "The bowl contains " . $this->contents . " soup!"; } TomTees 2 PART QUESTION (so I don't have to ask two questions 😀 )
1.)
2.) On my particular coding, I can use either the NULL/isset style, or, the ""/== style, with no errors or ill effects.... but something tells me the expert PHP coders prefer one over the other... Thank you!! Edited July 10 by ChenXiuHow can I change Yes to No for the NULL field in the mysql db structure. E.g. (Table name: address and Column name : state) <?php $link = mysql_connect("localhost","db","pw"); mysql_select_db("db"); mysql_query("alter table address modify state null NO); $result = mysql_query($query); $sent = "Edit Successful."; echo ($sent); ?> I'm trying to display the null values only, where am I going wrong?
SELECT IFNULL(trans, 0) AS 'Transactions', customer.cID AS 'Customer ID', customer.surname AS 'Surname' FROM customer LEFT JOIN account ON (account.cID = customer.cID) WHERE IFNULL(trans, 0); My code; $sql = "SELECT SUM(IF(`submitdate` IS NULL , 1 , 0 )) as 'Survey Started But Not Completed' FROM `survey_$surveyid`"; $statement = $dbh->prepare($sql); $statement->execute(); $result = $statement->fetch(PDO::FETCH_OBJ); //pass that data to an objectIs returning; stdClass Object ( [Survey Started But Not Completed] => ) I need to set this to a vaule that I can actually print out on the screen... Like a "0" for example. At the moment it is just a NULL. How do I do this?? These don't work; $result = 0; $result = array(['Survey Started But Not Completed'] => "0"); hi all, i have a script in where the user enters there post code and it will provide a price for delivery, how would i echo "please contact us for price" if the db record for the price is (null) or empty?? <?php mysql_connect ("localhost", "root","password") or die (mysql_error()); mysql_select_db ("postcode"); $term = $_POST['term']; $sql = mysql_query("SELECT * FROM uk_postcodes, zones WHERE postcode LIKE '%$term%' AND zone_id = zone_large "); while ($row = mysql_fetch_array($sql)){ echo '<br/> Postcode: '.$row['postcode']; echo '<br/> Town: '.$row['town']; echo '<br/> County: '.$row['county']; echo '<br/> Small: £'.$row['price_s']; echo '<br/> Medium: £'.$row['price_m']; echo '<br/> Large: £'.$row['price_l']; echo '<br/><br/>'; } ?> Hi... i have table EMP_NO LOGIN - DATETIME, NULL LOGOUT - DATETIME, NULL I have this query: Code: [Select] <?php include 'config.php'; $currentEmpID = $_SESSION['empID']; if(!isset($_POST['submit_'])){ $DATE1 = $_POST['firstinput']; $DATE2 = $_POST['secondinput']; $sql = "SELECT EMP_NO, LOGIN, LOGOUT FROM employee_attendance WHERE DATE(LOGIN) BETWEEN '$DATE1' AND '$DATE2'"; $attendance = $conn3->GetAll($sql); $smarty->assign('attendance', $attendance); } $smarty->display('header_att.tpl'); $smarty->display('empAttendance.tpl'); $smarty->display('footer.tpl'); ?> My problem is I have data EMP_NO 00000223 LOGIN NULL LGOUT NULL this data was not displayed because it has a null value.. i want it to displayed also and highlight it because it has a value null.. Thank you I'm trying to force a NULL value into the database with a hidden value and I can't seem to get it to work. Any suggestions? <input type="hidden" name="inactive" value=""/> I also tried here too: $sql="UPDATE product SET inactive='NULL' WHERE id = '$item_id'"; I have an array which can be empty; in this case, I get error. I want to skip when the array is empty. When calling the array from file, I use the following code if (file_exists($array_file)) { "the doing code" } else { echo "File does not exist"; } How I should use this If Else condition when array comes from a string like $test_array = explode(', ', $examplestring, -1); I have a for loop that I am looping through. When I use var_dump() I get 10 results (which is what I should be getting). However, when I loop through it using a for() loop I am getting an extra 11th value that is blank. What would be causing this? My code is below: Code: [Select] $legacy = new Legacy(); // Loop through each film in the shopping cart for ($i = 0; $i < $myShoppingCartInfo['row_count']; $i++) { // Determine if the film is of a legacy series if ($legacy->isLegacyCheckout($myShoppingCartInfo['data']['PRODUCTSKUID'][$i])) { // Declare array to hold all legacy film entity_id values $legacyEntityIds = array(); $legacyEntityIds = $legacy->getLegacyEntityIds($myShoppingCartInfo['data']['PRODUCTSKUID'][$i]); var_dump($legacyEntityIds); // Loop through all legacy video entity_id's for ($i = 0; $i < $legacyEntityIds['data']['ENTITY_ID']; $i++) { // Grant streaming access to each entity id echo $legacyEntityIds['data']['ENTITY_ID'][$i] . '<br />'; // Insert the product into the order items table /*$query = "INSERT INTO tblOrderItems (OrderItemID, ProductSKUID, RentalDate, OrderID) VALUES (##newID##, '".$myShoppingCartInfo['data']['PRODUCTSKUID'][$i]."', ".($myShoppingCartInfo['data']['RENTALDATE'][$i] === NULL ? 'NULL' : "'".$myShoppingCartInfo['data']['RENTALDATE'][$i]."'").", ".$newOrderID.")"; $cartItemInserted = counter_id_insert($query,'OrderItemID'); $curr_user->addNewEntityLicense($legacyEntityIds['data']['ENTITY_ID'][$i]);*/ } } When I view the source code of the output page in a browser, this is what I see: array(5) { ["row_count"]=> int(10) ["col_count"]=> int(1) ["col_names"]=> array(1) { => string(9) "entity_id" } ["col_types"]=> array(1) { => string(3) "int" } ["data"]=> array(1) { ["ENTITY_ID"]=> array(10) { => string(7) "1681799" [1]=> string(7) "1681872" [2]=> string(7) "1681871" [3]=> string(7) "1681870" [4]=> string(7) "1681869" [5]=> string(7) "1681868" [6]=> string(7) "1681867" [7]=> string(7) "1681866" [8]=> string(7) "1681865" [9]=> string(7) "1681864" } } } 1681799<br />1681872<br />1681871<br />1681870<br />1681869<br />1681868<br />1681867<br />1681866<br />1681865<br />1681864<br /><br /> |