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PHP - Query Returns Nothing
this is supposed to display the distinct first two characters of a string(billnmbr) in a combobox
ive tried this on phpmyadmin and it returns 08.. which is correct when i run this on my server, it returns nothing.. there must be something wrong with the way i called the billnmbr does anybody have an idea on how i could call and display this correctly? thanks and have a nice day. <?php $mquery = "select distinct(substring(`billnmbr`, 1, 2)) from `tbltest`"; $mres =mysql_query($mquery) or die(mysql_error()); while($mrow=mysql_fetch_array($mres)){ $mm = $mrow['billnmbr']; echo "<option value=" . $mrow['billnmbr'] . ">"; echo $mrow['billnmbr'] . "</option>"; } echo " </select>"; ?> Similar TutorialsHello. This one is driving me up the wall. It used to work, I have a sports league database (MsSql) and one of the tables contains team Captain's first name, last name, etc.. To create a unique but simple login, the query searches for the captain's first initial of their first name + lastname + league. However since there is the remote possibility of two similar usernames (i.e. David Smith) if the last name is found, the query adds an integer to the end of the last name. So the result would look like this: DSmithM4A At the moment, my database is completely blank. However what is happening when I run the query below is I get: DSmith1M4A Can someone figure out why this code is not working correctly? If there is not more than one "Smith" no integer is supposed to be returned. There is something in the "Captain Count" that I think is not working correctly. Code: [Select] /* * Gather captian information */ $strCaptainFirstInitial = substr($arrPost['cpt_first'],0,1); $strCaptainLastName = preg_replace("/[^a-zA-Z]/","",$arrPost['cpt_last']); $strCaptainName = strtolower($strCaptainFirstInitial.$strCaptainLastName); $sqlCaptainSearch = sprintf ( $strCaptainSearch, $strCaptainName ); $resCaptainSearch = mssql_query($sqlCaptainSearch,$conDB); /* * Create unique captain login and Password */ $strCaptainCount = (mssql_num_rows($resCaptainSearch) > 0) ? mssql_num_rows($resCaptainSearch) : "" ; $strLeagueTypeInitial = substr($arrLeagueSearch['type'],0,1); $strCaptainLogin = $strCaptainName . $strCaptainCount . $strLeagueTypeInitial . $arrLeagueSearch['size'] . $arrLeagueSearch['division']; Hi all ! The following code has two similar queries. Query 1 and Query 2. While Query 1 works just fine, Query 2 simply fails persistently. I have not been able to figure out why?
I have checked every single variable and field names but found no discrepancy of any sort or any mismatch of spelling with the fields in the database.
I would be grateful if anybody can point out what is preventing the second query from returning a valid mysqli object as the first one does. Here is the code for the two.
<?php
$db_host = "localhost";
$db_user ="root";
$db_pass ="";
$db_database ="allscores";
//////////////// CHECKING VARIABLE FILEDS IN A UPDATE QUERY //////////////////////////////////
//////////////// QUERY 1 /////////////////////////////////////////////////////////////////////
/*
$db_database ="test";
$db_table ="users";
$RecNo = 1;
$field1 = 'name';
$field2 = 'password';
$field3 = 'email';
$field4 = 'id';
$val1 = "Ajay";
$val2 = "howzatt";
$val3 = "me@mymail.com";
$con = mysqli_connect($db_host,$db_user,$db_pass,$db_database) or die('Unable to establish a DB connection');
$query = "UPDATE $db_table SET $field1 = ?,
$field2 = ?, $field3 = ?
WHERE $field4 = ? "; // session terminated by setting the sessionstatus as 1
$stmt = $con->prepare($query);
var_dump($stmt);
$stmt->bind_param('sssi',$val1,$val2,$val3,$RecNo);
if($stmt->execute())
{
$count = $stmt->affected_rows;
echo $count;
}
//////////////// QUERY 1 END /////////////////////////////////////////////////////////////////////
*/
//////////////// QUERY 2 /////////////////////////////////////////////////////////////////////
$con = mysqli_connect($db_host,$db_user,$db_pass,$db_database) or die('Unable to establish a DB connection');
$table = 'scores';
$date = date('Y-m-d H:i:s');
/*
$prestr = "Mrt_M_";
$STATUS = $prestr."Status";
$S_R1 = $prestr."Scr_1";
$S_R2 = $prestr."Scr_2";
$PCT = $prestr."PPT";
$DPM = $prestr."DSP";
$TIMETAKEN = $prestr."TimeTaken";
*/
$STATUS = "Mrt_M_Status";
$S_R1 = "Mrt_M_Scr_1";
$S_R2 = "Mrt_M_Scr_2";
$PPT = "Mrt_M_PPT";
$DSP = "Mrt_M_DSP";
$TIMETAKEN = "Mrt_M_TimeTaken";
$TimeOfLogin = $date;
$no_of_logins = 10;
$time_of_nLogin = $date;
$m_scr_row1 = 5;
$m_scr_row2 = 5;
$m_ppt = 20;
$m_dsp = 60;
$m_time = 120;
$date = $date;
$RecNo = 24;
$query = "UPDATE $table SET
TimeOfLogin = ?,
no_of_logins = ?,
time_of_nLogin = ?,
$S_R1 = ?,
$S_R2 = ?,
$PPT = ?,
$DSP = ?,
$TIMETAKEN = ?,
$STATUS = '1',
TimeOfLogout = ?,
WHERE RecNo = ?";
$stmt = $con->prepare($query);
var_dump($stmt);
$stmt->bind_param('sisiiddssi',$TimeOfLogin,$no_of_logins,$time_of_nLogin,$m_scr_row1
$m_scr_row2,$m_ppt,$m_dsp,$m_time,$date,$RecNo);
if($stmt->execute()) echo " DONE !";
?>
Thanks to all
I have a table korisnici in SQLite with INTEGER field aktivan that can have only 0 or 1 value (CHECK constraint). Field aktivan has value 0 but PHP returns value 1, why? Is this a bug? This is PHP code that I am running:
$sql = "SELECT ime, aktivan FROM korisnici WHERE lower(ime) = '" . $ime . "'" . " AND sifra = '" . $_POST["sifra"] . "'";
$result = $db->query($sql);
$row = $result->fetchArray(SQLITE3_ASSOC);
$row['aktivan'] = 1 but in table the value is 0. When I run same query in DB Browser for SQLite I get correct value 0. Is this a bug? soo i get info from a form and i want to search for it in my db .. if i find user_name AND user_password the $result returns true .. otherwise false .. <?php require_once("../includes/connection.php"); ?> <?php $user_name=$_POST["user_name"]; $user_password=$_POST["user_password"]; ?> <?php $query = "SELECT * FROM users WHERE '{$user_name}' = user_name AND '{$user_password}'= user_password "; $result = mysql_query($query,$connection); if($result==false){ redirect("../index.php");} elseif($result==true){ redirect("../welcome.php"); } ?> <?php //close connection! mysql_close($connection); ?> but the problem is that it always returns TRUE !!!! whats the problem here exactly ??? Hi, I have a sql query that executes perfectly in phpMyAdmin but when applied to a PHP script returns nothing. It won't even start a "while" loop. Please review code snippits below. Submission Script calls function in Database Class: $result = $database->testReturnId($_POST['years'], $_POST['model']); while($row = mysql_fetch_array($result)) { $i = $i + 1; echo "ROW " . $i . " " . $row['auto_id'] . " " . $row['auto_year'] . " " . $row['auto_year_high'] . " " . $row['auto_make'] . " " . $row['auto_model'] . "<br />"; } Database Class Function: function testReturnId($year, $model){ global $form; $q = 'SELECT * FROM tbl_svc_auto WHERE auto_model = "$model" AND "$year" BETWEEN auto_year AND auto_year_high'; if(!mysql_query($q, $this->connection)){ return false; }else{ return $result = mysql_query($q, $this->connection); } } I have moved the query to the submission script and replaced the variables with actual data and it still will not work. I can take the same query with out the variables and input it into phpMyAdmin and it works great. On the submission script, I added a row number to the echo thinking that it would a least return the word "ROW" and it does not. I have tested each step of the execution of the query and have not returned any errors. I do not receive any errors at all even during execution. It just seems to refuse to run the "while" statement. I do not know where my problem is......Please help. Thank you in advance for any assitance you can offer. Joshua I'm trying to write a php page that displays data from a JOIN query for a specific ID table view brandinfo ID, brand, discounttype 1, antioni, no discount brandproducts brandID, producttype, price 1, Tshirt, 20.00 1, Pants, 30.00 1, Shoe, 40.00 the returned result is 1 antioni, no discount, Tshirt, 20.00, 2 antioni, no discount, Pants, 30.00 3 antioni, no discount, Shoe 40.00 The way I want the page to be displayed is ------------------ Antioni (at the top) Table 1. Tshirt 20.00 2. Pants 30.00 3. Shoe 40.00 no discount (at the bottom) ---------------------------- How should I construct the PHP page from the result since they're retrieved as rows? Hello, i have a query that returns the rows of certain parts of a databse. i made a button on the end of each echo, to make sure that, the button being printed with each echo, correlates to that echo. However, the SESSIONS only return the same varaibles each time. The last in the list.
$sql = "SELECT p.Title, p.PerfDate, p.PerfTime FROM performance AS p INNER JOIN production AS r ON p.Title = r.Title";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo '<tr>
<td>'.$row['Title'].'</td>
<td>'.$row['PerfDate'].'</td>
<td>'.$row['PerfTime'].'</td>
<td><button name="availability" value="'.$row['Title'].'">Availability</button></td>
</tr>';
$_SESSION["Title"] = $row['Title'];
$_SESSION["PerfDate"] = $row['PerfDate'];
$_SESSION["PerfTime"] = $row['PerfTime'];
}
} else {
echo "0 results";
}
$_SESSION["name"] = $_GET["name"];
?>
How am i able to make it so each button correlates to each row?
There's a screenshot of how it looks for reference Edited December 8, 2019 by AdamSteele I'm getting frusted over an if else statememt that won't work. It is calling one variable from the table. If the variable is in the table it will display one thing, if the variable is absent, it should display something else. I've tried !empty, isset, even $num_rows =1, but the content in the else doesn't display. Here's the code: Code: [Select] <? $result = mysql_query( "SELECT * FROM codeWords WHERE Mycode = '$MyCode' " ) or die("SELECT Error: ".mysql_error()); $num_rows = mysql_num_rows($result); while ($row = mysql_fetch_array($result)) { extract($row); if (!empty( $MyCode )) { echo "here is the code word"; } else { echo "you need to go back and get the code"; } $row_count++; } mysql_close; ?> When the code variable is correct the "here is the code word" displays. when the code variable is incorrect, what should appear as "you need to go back and get the code" is just blank. Any thoughts on how to structure the query in such a way that eh incorrect text appears? Thanks, im not new enough for this to take so long to figure out, but for some reason $rs = mysql_query("SELECT * FROM cl"); returns the last record instead of all the records in the table. anyone know why it wouldnt take everything? I'm hoping someone can help me cos my hair's going white with this one. I'm trying to put together a script that acts as web-based interface to an SQL server. There are actually two parts, admin.htm and admin.php. The first part is just a form that passes login credentials to the PHP file. That part seems to work fine, but I'll post the source anyway: - Code: [Select] <!DOCTYPE HTML> <html lang="en"> <head> <title>SQL admin login</title> <meta charset="iso-8859-1" /> </head> <body> <form action="admin.php" method="post"> <label for="username">Username: -</label> <br /> <input type="text" name="username" id="username" /> <br /> <br /> <label for="password">Password: -</label> <br /> <input type="password" name="password" id="password" /> <br /> <br /> <label for="server">Server: -</label> <br /> <input type="text" name="server" id="server" /> <br /> <br /> <label for="database">Database: -</label> <br /> <input type="text" name="database" id="database" /> <br /> <br /> <input type="submit" value="Login" /> <input type="reset" value="Reset" /> </form> </body> </html> Following is the content of admin.php. By this point I can see the connection in MySQL Workbench, and when I submit the query 'SELECT * FROM subscribers' it's being stored in '$_POST['query']', but 'mysql_query($_POST['query'],$_SESSION['con']);' is returning nothing. There is definitely a record in that table, and the user I'm logging on with has permission to run the 'SELECT' command against this database, so I can't figure out why mysql_query(); is returning nothing: - Code: [Select] <!DOCTYPE HTML> <?php session_start(); if(!$_SESSION['con']) { if(!($_POST['username'] || $_POST['password'])) { if(!($_SESSION['username'] || $_SESSION['password'])) { $error="Username and password variables empty."; } } else { $_SESSION['username']=mysql_real_escape_string($_POST['username']); $_SESSION['password']=mysql_real_escape_string($_POST['password']); $_SESSION['server']=mysql_real_escape_string($_POST['server']); $_SESSION['database']=mysql_real_escape_string($_POST['database']); $_SESSION['con']=mysql_pconnect($_SESSION['server'],$_SESSION['username'],$_SESSION['password']); if(!$_SESSION['con']) { $error="Failed to connect to server."; } else { $database=mysql_select_db($_SESSION['database'],$_SESSION['con']); if(!$database) { $error="Failed to connect to database."; } } } } if(!$_POST['query']) { $error="No query submitted."; } else { $result=mysql_query($_POST['query'],$_SESSION['con']); if(!$result) { $error="Query returned nothing."; } } ?> <html lang="en"> <head> <title>SQL admin interface</title> <meta charset="iso-8859-1" /> </head> <body> <form action="admin.php" method="post"> <textarea name="query" rows="10" cols="50">SELECT * FROM subscribers</textarea> <br /> <br /> <input type="submit" value="Submit query" /> </form> <?php if($error) { echo $_POST['query']."<br /><br />".$result."<br /><br />".$error; die(); } else { while($row=mysql_fetch_assoc($result)) { echo $row['name']." ".$row['email']; echo "<br />"; } } ?> </body> </html> Can anyone help? MOD EDIT: [code] . . . [/code] tags added. Hi again everyone! I'm sitting here, developing the same site - on my localhost the site is working great! Then when I upload it, to another testserver (production for developers), I get following: Notice: Undefined variable: db in C:\xampp\htdocs\lucas\sidebar.php on line 3 Fatal error: Call to a member function prepare() on a non-object in C:\xampp\htdocs\lucas\inc\functions.php on line 13 I'm wondering why I get it, because if I do a try / catch on the db connection, and do a select * on the content table, it returns results, the paths should be almost the same.. On localhost I run: localhost/lucas/, on the prod-server I run liveip/lucas/ Can any spot the problem? I'm not a big xampp dude myself, I've attached an dump of my PHP info also, just rename the .php to .html Thank you a lot! Lucas R / Zerpex Hi i am running a wamp server on windows 7 i am trying to php exec command to run a command but the output returns 1 does this mean the cmd failed to execute ? Code: [Select] $openssl_cmd = "($OPENSSL mime -sign -signer $MY_CERT_FILE -inkey $MY_KEY_FILE ". "-outform der -nodetach -binary <<_EOF_\n$data\n_EOF_\n) | " . "$OPENSSL smime"." -encrypt -des3 -binary -outform pem $PAYPAL_CERT_FILE"; <?php if('00' == '000') echo 'weired'; ?> Please consider the code above. Strangely it echoes the word "weired". Also if('00' == '000000000000000') returns true and so forth. What is really going on? Thanks in advance By the way the php version is 5.3.9 on wampserver 2.2 Hi, I am trying to verify if the given url exists or not, by using file_exists() function. It always returns 'FALSE' , according to my understanding it happening because the file to be checked on the given url is located in safe mode. Could anyone please suggest as to how this could be overcome, by similar function or by using alternative method. Regards Abhishek Madhani When you do a search at (right sidebar): http://ourneighborhooddirectory.org the $_GET result is always blank. The site is hosted at Bluehost and this is a Wordpress site. You can see the result of the query because I write them out. Here is the form code: Code: [Select] <form method="get" action="http://ourneighborhooddirectory.org/search-results/" > Search the Directory: <input type="text" name="srch"/> <input type="submit" value="Search" /> </form> Here is the PHP code on the action page: <?php $srchVal = $_GET["srch"]; echo $_GET['srch']; echo "------------ ". $srchVal." ------------"; print_r($_GET); echo $_SERVER['QUERY_STRING']; ?> Any ideas on why $_POST and $_GET always return blank? Thanks for the help... I have a table called users with a fieldname called service_id. In a table called services I have id and name. I want to query the users table and, based on the service_id, display the name of the service (which is stored in the services table). However, when I try this code I get No records returned. I later echo out under a while statement $row['name']; or $row['id'] $query = "SELECT users.username, users.lname, users.fname, users.service_id, services.name, services.id FROM users, services WHERE users.inst_id = '".$userarray['inst_id']."' and users.id !='".$userarray['id']."' and users.service_id = services.id "; I am trying to unserialize this string: a:3:{s:3:\"zip\";s:5:\"55068\";s:4:\"city\";s:9:\"Rosemount\";s:5:\"state\";s:2:\"MN\";} but I unserialize is returning false. Why is it doing that? $info = unserialize($_COOKIE['zipcode']); var_dump($info); Hi, I am trying to expand my very limited PHP by building a CMS. It's all new to me so please be courteous with your replies ? I little history of my problem. I am running xampp on a windows 10 running php 5.6.38
I have taken two video courses and on each I got stumped halfway through with a very similar problem. I am copying everything word for word and even downloaded the course file on the first one and still got the same problem so I'm wondering if there is something wrong with my setup - config?.
The problem.
This is the code:
<?php
error_reporting(E_ALL);
ini_set('display_errors',1);
include("includes/database.php");
include("includes/header.php");
if ($connection) {
echo "Great, we're connected to the database<br><br>";
}else {
echo "Bummer, it didn't connect to the database!";
}
$query = "SELECT * FROM categories";
$result_set = mysqli_query($connection, $query) ;
if (!$result_set) {
die("Query Failed".msqli_error($connection));
}
while ($row = mysqli_fetch_array($result_set)) {
echo $row['cat_title'];
}
var_dump($result_set);
?>
Great, we're connected to the database
object(mysqli_result)#3 (5) { ["current_field"]=> int(0) ["field_count"]=> int(2) ["lengths"]=> NULL ["num_rows"]=> int(0) ["type"]=> int(0) }
Question:
Thanks for your help. Hi All, I am pretty new to PHP ( I am a C Programmer). I am trying a simple shopping cart website for my friend. IN one of the pages, I am submitting some info, which is actually not showing up in my server side php. Below is the HTML code to submit. <a href="cart.php?action=add&p=5&size=small"> <img src="images/add_to_cart.gif"/> </a> I tried to check the values of the parameters using $action = $_GET['action'] ; Action stored empty. Then I checked the whole array print_r($_GET) which returned Array( ) So, for some reason the value is always empty. I tried the declaration GLOBAL $_GET, but no use. Any help/directions - much appreciated! |
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