PHP - Mysql And Php Help With Variable
Hi,
I am very new to this. Can someone point me in the right direction please. I have mysql database which contains a variable $pricelist. The client is registered via an admin area. Is sent the login (email and password). Once logged in they see their details. All this works fine. Each client will have a different price list, total number of price lists options will be about 10. In the database the variable will return pricelist1.pdf or pricelist2.pdf etc. This is also OK. My question is how do I provide a hyperlink to the pricelist1.pdf file (which is located on the server) for the client to view and download if they wish? Thanks in advance. Similar TutorialsI am trying to allow the user to update a variable he chooses by radio buttons, which they will then input text into a box, and submit, to change some attributes. I really need some help here. It works just fine until I add the second layer of variables on top of it, and I can't find the answer to this question anywhere. <?PHP require('connect.php'); ?> <form action ='' method='post'> <select name="id"> <?php $extract = mysql_query("SELECT * FROM cars"); while($row=mysql_fetch_assoc($extract)){ $id = $row['id']; $make= $row['make']; $model= $row['model']; $year= $row['year']; $color= $row['color']; echo "<option value=$id>$color $year $make $model</option> ";}?> </select> Which attribute would you like to change?<br /> <input type="radio" name="getchanged" value="make"/>Make<br /> <input type="radio" name="getchanged" value="model"/>Model<br /> <input type="radio" name="getchanged" value="year" />Year<br /> <input type="radio" name="getchanged" value="color" />Color<br /><br /> <br /><input type='text' value='' name='tochange'> <input type='submit' value='Change' name='submit'> </form> //This is where I need help... <?PHP if(isset($_POST['submit'])&&($_POST['tochange'])){ mysql_query(" UPDATE cars SET '$_POST[getchanged]'='$_POST[tochange]' where id = '$_POST[id]' ");}?> I am trying to extract a value that is stored in a variable from my database. It will not work with how I have the code written. If I replace .$tournament with what I want it to find, it will give me the correct result. ANy suggestions? Below is my code: Code: [Select] $result = mysql_query("SELECT id FROM `2012_tournaments` WHERE tournament = .$tournament"); while($row = mysql_fetch_array($result)) { echo $row['id']; }; hello can anyone help me how to insert GET variable into mysql <?php $page_title = 'Personal Wellness'; include ('template/header.inc'); include_once('config.php'); $id = $_GET['id']; if(isset($_POST['submit'])) //if submit was pressed { if(strlen($_POST['height'])<1) //if there was no height { print "You did not enter a height."; } else if(strlen($_POST['weight'])<1) //no weight { print "You did not enter a weight."; } else if(strlen($_POST['bodyfat'])<1) //no bodyfat { print "You did not enter a Body Fat Range"; } else if (strlen ($_POST['bodywater'])<1) //no bodywater { print "You did not enter a Body Water Range"; } else if( strlen($_POST['musclemass'])<1) //no musclemass { print "You did not enter a Muscle Mass"; } else if (strlen ($_POST['physiqueratt'])<1) //no physiqueratt { print "You did not enter a Physique Ratings"; } else if (strlen ($_POST['bonemass'])<1) //no bonemass { print "You did not enter a Bone Mass"; } else if (strlen ($_POST['bmr'])<1) //no bmr { print "You did not enter a BMR"; } else if (strlen ($_POST['basalmetabolic'])<1) //no basalmetabolic { print "You did not enter a Basal Metabolic Age"; } else if (strlen ($_POST['visceralfat'])<1) //no visceralfat { print "You did not enter a Visceral Fat"; } else if(strlen($_POST['registrationmonth'] && $_POST['registrationday'] && $_POST['registrationyear'])<1) // no date { print "You did not enter a date of birth"; } else //all fields met { $id=$_GET['id']; $height=$_POST['height']; $weight=$_POST['weight']; $bodyfat=$_POST['bodyfat']; $bodywater=$_POST['bodywater']; $musclemass=$_POST['musclemass']; $physiqueratt=$_POST['physiqueratt']; $bonemass=$_POST['bonemass']; $bmr=$_POST['bmr']; $basalmetabolic=$_POST['basalmetabolic']; $visceralfat=$_POST['visceralfat']; $date=$_POST['registrationyear'] . '-' . $_POST['registrationmonth'] . '-' . $_POST['registrationday']; $insertadmin="INSERT into personalwelness (m_id,height,weight,body_fat,body_water,muscle_mas s,physique_ratt,bone_mass,bmr,basal_metabolic,visc eral_fat,evaluation_date) values ('$id','$height','$weight','$bodyfat','$bodywater','$mus clemass','$physiqueratt','$bonemass','$bmr','$basa lmetabolic','$visceralfat','$date')"; //registering admin in databae echo $insertadmin; $insertadmin2=mysql_query($insertadmin) or die("Could not insert admin"); print "Personal Wellness Successfully Submitted"; } } ?> <form method="post" class="form" action="<?php echo $_SERVER['PHP_SELF'];?>"> <fieldset><legend>Enter Personal Wellness Information in the form below:</legend> <table width="80%" border="0"> <tr> <td width="16%">Height(CM)</td> <td width="2%">:</td> <td width="82%"><label for="height"></label> <input type="text" name="height" id="height" value="<?php if (isset($_POST['height'])) echo $_POST['height'];?>" /></td> </tr> <tr> <td>Weight(KG)</td> <td>:</td> <td><label for="weight"></label> <input type="text" name="weight" id="weight" value="<?php if (isset($_POST['weight'])) echo $_POST['weight'];?>" /></td> </tr> <tr> <td >Body Fat Range</td> <td>:</td> <td><label for="body fat"></label> <input type="text" name="bodyfat" id="bodyfat" value="<?php if (isset($_POST['bodyfat'])) echo $_POST['bodyfat'];?>" ></td> </tr> <tr> <td>Body Water Range(%)</td> <td>:</td> <td><label for="bodywater"></label> <input type="text" name="bodywater" id="bodywater" value="<?php if (isset($_POST['bodywater'])) echo $_POST['bodywater'];?>"/></td> </tr> <tr> <td>Muscle Mass</td> <td>:</td> <td><label for="musclemass"></label> <input type="text" name="musclemass" id="musclemass" value="<?php if (isset($_POST['musclemass'])) echo $_POST['musclemass'];?>"></td> </tr> <tr> <td>Physique Ratings</td> <td>:</td> <td><label for="physiqueratt"></label> <input type="text" name="physiqueratt" id="physiqueratt" value="<?php if (isset($_POST['physiqueratt'])) echo $_POST['physiqueratt'];?>"></td> </tr> <tr> <td>Bone Mass</td> <td>:</td> <td><label for="bonemass"></label> <input type="text" name="bonemass" id="bonemass" value="<?php if (isset($_POST['bonemass'])) echo $_POST['bonemass'];?>" /></td> </tr> <tr> <td>BMR</td> <td>:</td> <td><label for="bmr"></label> <input type="text" name="bmr" id="bmr" value="<?php if (isset($_POST['bmr'])) echo $_POST['bmr'];?>"/></td> </tr> <tr> <td>Basal Metabolic Age</td> <td>:</td> <td><label for="basalmetabolic"></label> <input type="text" name="basalmetabolic" id="basalmetabolic" value="<?php if (isset($_POST['basalmetabolic'])) echo $_POST['basalmetabolic'];?>"></td> </tr> <tr> <td>Visceral Fat</td> <td>:</td> <td><label for="visceralfat"></label> <input type="text" name="visceralfat" id="visceralfat" value="<?php if (isset($_POST['visceralfat'])) echo $_POST['visceralfat'];?>"></td> </tr> <tr> <td>Evaluation Date</td> <td>:</td> <td> <?php echo date_picker("registration")?></td> </tr> </table> </fieldset> <div align="center"><input type="submit" name="submit" value="Submit" /> </div> </form> <?php function date_picker($name, $startyear=NULL, $endyear=NULL) { if($startyear==NULL) $startyear = date("Y")-100; if($endyear==NULL) $endyear=date("Y")+50; $months=array('','January','February','March','Apr il','May', 'June','July','August', 'September','October','November','December'); // Month dropdown $html="<select name=\"".$name."month\">"; for($i=1;$i<=12;$i++) { $html.="<option value='$i'>$months[$i]</option>"; } $html.="</select> "; // Day dropdown $html.="<select name=\"".$name."day\">"; for($i=1;$i<=31;$i++) { $html.="<option $selected value='$i'>$i</option>"; } $html.="</select> "; // Year dropdown $html.="<select name=\"".$name."year\">"; for($i=$startyear;$i<=$endyear;$i++) { $html.="<option value='$i'>$i</option>"; } $html.="</select> "; return $html; } ?> <?php include ('template/footer.inc'); ?> Hi again everyone, I am trying to compare a php variable to the value of a row in mysql, but keep getting the following error: Code: [Select] Wrong perameter count for mysql_result() What syntax would I use to accomplish this? Here is my code: // connects to server and selects database. include ("../includes/dbconnect.inc.php"); // table name $table_name = "availability"; // split up date into 3 separate fields $date = "12/25/2010"; $date_split = explode("/", $date); $month = $date_split[0]; $day = $date_split[1]; $year = $date_split[2]; // check if earth room is already reserved $taken = "Reserved"; $sql = "SELECT earth_room FROM $table_name WHERE month='$month' AND day='$day' AND year='$year' LIMIT 1"; $result = mysql_query($sql) or trigger_error("A mysql error has occurred!"); $row = mysql_result($result); if($row == $taken){ echo "Room Already Reserved"; } else{ echo "Room Available"; } Thanks for the help, kaiman You guys are great, thanks again for the help last week. Now I almost got this working but a small hiccup. here is my code: Code: [Select] <?php include("config.php"); $my_t=getdate(date("U")); $my_t1=$my_t[weekday]; $result = mysql_query("SELECT * FROM tourney where day_of_week = '$my_t1'") or die(mysql_error()); if ($result == '') echo "<br>Empty Set\n"; print_r ($my_t1); This prints correctly Code: [Select] while ($result = mysql_fetch_array($result,MYSQL_ASSOC)) { This is my error. "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /hermes/bosweb/web017/b172/ipg.dswdesignsnet/pp4c/charity1.php on line 8" Code: [Select] print "<b>Starting Time: <br></b>".$row{'start_time'}."<br><b>Tournament: </b><br>".$row{'tourney'}."<br><b>Buy-in: <br>".$row{'buy_in'}."<br><b>Starting Chips: <br>".$row{'start_chips'}."</font><p>"; This prints headers correctly, but no variables. Code: [Select] } mysql_close($dbh); ?> What did I forget to do or what did I do wrong. I'm still learning mysql and php. Hi im trying to set that if a certain statement is echoed then set variables to an amount but they dont seem to be inserting into the table Code: [Select] if ($match_qutoe == "".$username2." lands a uppercut"){ $points = "4"; $fighter = $username2; } $query = mysql_query("INSERT INTO fight (bluecornerusername, redcornerusername, round, fighter, points, statement) VALUES('$username', '$username2', '$round', '$fighter', '$points', '$match_quote')"); its only $points and $fighter that are not inserting Thanks Hi anyone here know how to insert $GET variable into mysql, i don't know how to put this variable between curly bracket, when i put on top insert query, i got error 'Could not insert admin'...please help Code: [Select] <?php mysql_connect("localhost","root","") or die(mysql_error()); mysql_select_db("healthsystem") or die(msql_error()); [color=red]//GET varibable $id = $_GET['id'];[/color] // file properties $file = $_FILES['image']['tmp_name']; if (!isset($file)) echo "Please select an image"; else { $image = addslashes(file_get_contents($_FILES['image']['tmp_name'])); $image_name = addslashes($_FILES['image']['name']); $image_size = getimagesize($_FILES['image']['tmp_name']); if ($image_size==FALSE) echo "That's not an image."; else { $insert = "INSERT INTO image_tbl(m_id,name,image) VALUES ('$id','$image_name','$image')"; $insert2=mysql_query($insert) or die("Could not insert admin"); print "Personal Wellness Successfully Submitted"; } } ?> I have a mysql field that I want to store a php variable in and then retrieve it.
Example:
Color table has the field named colorBackground with the value of #FF0000
CSS table has the field named cssBackground with the value of .background { background-color: #444444; }
What I want to do is make Color/colorBackground a variable {{$backgroundcolor = rsColor['colorBackground']
Then I want to change the CSS/cssBackground value to .background { background-color:$backgroundcolor; }
Creating the $backgroundcolor works fine. I'm just not sure how to put the $backgroundcolor in my CSS/cssBackground field.
Can anyone point out how to write a MySQL query with a PHP variable in the WHERE clause. I've tried {} {'xx'} and () and it still doesn't work. Here is the code <?php ini_set('display_errors',1); error_reporting(E_ALL|E_STRICT); include ("include/connect.php"); include ("include/session.php"); $username = $session->userinfo['username']; $result = mysql_query("SELECT email FROM customer WHERE user = {'$username'} "); while($row = mysql_fetch_array($result)) { $custemail = $row['email']; } echo "Session username: " . $username . ""; echo "Session customer email: " . $custemail . ""; ?> So I'm trying to show the email address for a record that matches the username of the user logged in. I really appreciate the help. hey all so I have this bit down: Code: [Select] $query="SELECT `2010 Region Code` AS codes FROM locations"; $results = mysql_query($query); $options=""; $options = "<select location='codes'>"; while($nt=mysql_fetch_assoc($results)) { $thing=$nt["codes"]; $options.="\r\n<option value ='{$nt['codes']}'> {$nt['codes']}</option>"; } $options .="\r\n</select>"; echo $options; what I'm trying to do is grab the selection from the drop down and display it as a table (the sql query would be extended should we manage to figure this one out I've tried Code: [Select] echo"<form name='LOCATIONS' action='".$_SERVER['PHP_SELF']."' target='iframe' method='post'>"; any ideas? Hello, I am working on a project and I cant seem to get this to work. I have a database (Kingsbury) that stores a contact list. I am trying to make an page were a user can search the database by state. The code from the HTML site is: Code: [Select] <form action = "sqlllist.php" method="post"><br /> Please enter the state you would like to search: <input type="text" name="state" /><br /> <p><input type="submit" value="Enter" /></p> </form> And then from the .php file Code: [Select] <?php $DBConnect = mysql_connect("*****", "*****", "********"); if ($DBConnect===FALSE) echo "<p>Connection Failed!.</p>\n"; else { $State == $_POST['state']; $Result = mysql_select_db("Kingsbury", $DBConnect); $queryresult = @mysql_query("SELECT * FROM contacts WHERE State = $State'", $DBConnect); echo "<table width='100%' border='1'>\n"; echo "<tr><th>User ID</th> <th>First Name</th> <th>Last Name</th> <th>Address</th> <th>State</th> <th>City</th> <th>Zip Code</th> <th>Area Code</th> <th>Phone Number</th>\n"; While (($Row = mysql_fetch_row($queryresult)) !== FALSE) { echo "<tr><td>{$Row[0]}</td>"; echo "<td>{$Row[1]}</td>"; echo "<td>{$Row[2]}</td>"; echo "<td>{$Row[3]}</td>"; echo "<td>{$Row[4]}</td>"; echo "<td>{$Row[5]}</td>"; echo "<td>{$Row[6]}</td>"; echo "<td>{$Row[7]}</td>"; echo "<td>{$Row[8]}</td>"; } echo "</table>\n"; mysql_free_result($queryresult); mysql_close($DBConnect); } ?> This, to me seems like it should be working, but it outputs the entire database. I am working on a room availability calendar that has links for each day. When you click on a link it passes the day month and year in the URL to another page like this: echo "<td class=\"today\"> <a href=\"status.php?month=$month&day=$day_num&year=$year\">$day_num</a> </td>\n"; I can see that the correct information is being passed through the URL like this: Code: [Select] .../status.php?month=12&day=9&year=2010 Then the information is supposed to be passed to the MySQL query, but here is my question: How do I do this? I have a DB table set up, but the query is currently returning a blank page. Here is my current query: // connects to server and selects database. include ("../includes/dbconnect.inc.php"); // table name $table_name = "availability"; // query database for events $result = mysql_query ("SELECT id FROM $table_name WHERE month=$month AND year=$year AND day=$day_num LIMIT 1") or die(); if (mysql_num_rows($result) > 0 ) { while($row = mysql_fetch_array($result)) { extract($row); echo "<h1>Current availability for ".$row['month'] . "/" . $row['day'] . "/" . $row['year'] . "</h1>"; echo " <ul>"; echo " <li>Earth Room: " . $row['earth_room'] . "</li>"; echo " <li>Air Room: " . $row['air_room'] . "</li>"; echo " <li>Fire Room: " . $row['fire_room'] . "</li>"; echo " <li>Water Room: " . $row['water_room'] . "</li>"; echo " </ul>"; } } else { echo " <ul>"; echo " <li>Currently no reservations.</li>"; echo " </ul>"; } Any help is appreciated. Thanks, kaiman I have a form passing user's search term, ie name. $pname = mysql_real_escape_string($pname; Not sure on proper syntax to include this in my select statement. I also want it to return for partial string matched, like 'mik' for 'mike': $result = mysql_query( "SELECT * FROM guest WHERE name like '$pname'" ); Hey, I'm trying to save a MySQL-Link source as an instance variable in my query class. The problem is that MySQL sees it as an invalid MySQL-Link (supplied argument is not a valid MySQL-Link resource). What I'm I doing wrong? //Walle Code: [Select] private $connection; //Construct public function __construct(&$connection) { $this->connection = $connection; } //Where the error occurs public function exe_query($sql) { $result = mysql_query($sql, $this->connection); if(!$result) { throw new Exception(mysql_error()); } return $result; } Hello, I'm having some issues with PHP thinking that the variables that I send it are the actual columns in my database. First, I pull off Quadratic_Functions and introductory_problem from http://localhost:8888/algebra_book/Chapters/Quadratic_Functions/introductory_problem.php using the code below: Code: [Select] $chapter_page = $_SERVER['PHP_SELF']; $chapter_page_array = explode('/',$chapter_page); $size =count($chapter_page_array); $chapter = $chapter_page_array[$size-2]; $page = $chapter_page_array[$size-1]; $page_array = explode('.', $page); $page = $page_array[0]; Based on my printing of the variables $chapter and $page I think that it's doing what I want it to do. I then use the following function: Code: [Select] $supplemental_id = getSupplementalId($dbRead,$chapter,$page); to check out if the there's a supplemental_id for the Quadratic_Function chapter and introductory_problem page name via: Code: [Select] function getSupplementalId($read,$user_id,$chapter,$page) { $sql = "SELECT supplemental_id FROM instructors_supplemental WHERE page_name = $page AND chapter_name='$chapter"; return $read->fetchRow($sql); } If I stick in actual values, as seen below, the thing runs fine. Code: [Select] $sql = "SELECT supplemental_id FROM instructors_supplemental WHERE page_name = 'introductory_problem' AND chapter_name='Quadratic_Functions'"; But if I run it in the abstract version, with variables for page and chapter name (the first version), I get Fatal error: Uncaught exception 'PDOException' with message 'SQLSTATE[42S22]: Column not found: 1054 Unknown column 'introductory_problem' in 'where clause'' in... It's almost as if it thinks that my variables are the names of the columns. Any thoughts would be appreciated.... Hello Everyone, I have a quick question for you all, I think its fairly simple... I have created a database and I am using PHP to grab the data: $usera = $_SESSION['username']; $query2 = "SELECT * FROM tracker WHERE id = '$usera', hidden = yes"; mysql_query($query2) or die('Error, query failed : ' . mysql_error()); This hopefully will return multiple rows which look like this in the database. id username date reps hidden 1 supremebeing 2011-01-02 30 yes 4 supremebeing 2011-04-02 46 yes How would i turn each result into a variable eg: $date1 = 2011-01-02; $date2 = 2011-04-02; $reps1 = 30; $reps2 = 46; I think i have explained that well enough for you to understand, please reply if not though and i will provide more information. Thanks in Advance Hi I'm having a problem getting a query to work. I have a simple form with user input for start and end date with format: 2009-03-19 (todays date): $Startdate = $_POST['date']; This works well when something is entered into the form, and afterwards using my query: SELECT COUNT(*) as total FROM mydb WHERE Date BETWEEN '$Startdate' AND '$EndDate' ........ Problem is if user submits the form without entering anything in the date input fields, which makes sense. I want to check if inputs has been made, and if not set af default date, but can't make it work: if (isset($_POST['date']) && $_POST['date'] !='') { $Startdate = $_POST['date'];} else { $Startdate = '1980-01-01';} How can I set $Startdate to something that can be used in the query as below doesn't work? Please take a look at the following code (this would work if the statement was SELECT instead of UPDATE) But I need to grab the following: Code: [Select] $result = mysql_query("UPDATE `AUCTIONS_MAIN` SET Current_Price = Current_Price + '$increase_price', Last_Bidder_Time = '$last_bidder_time' WHERE PId = '$id'"); //grab the new updated current price to save on different bidding history DB while($row=mysql_fetch_array($result)) { $current_price = $row['Current_Price']; } As you can see, I need to give the new Current_Price value to the $current_price variable. Obviously the while function is incorrect, I would really appreciate any ideas and/or suggestions on how I can grab this value. PS: I can't make a new query call with a select statement in order to know the new Current_Price. I need the direct updated value coming from the Update Statement, is this possible? Thanks a lot in advance!! Hi Members,
I am search for the reason for the problem why my mysql query cannot fetch data and store in file based on id in $variable form. For example, $sql="SELECT * FROM mytable WHERE mine_id='1234'"; works for me. But when i use $sql="SELECT * FROM mytable WHERE mine_id='$id'";, files are created as empty. I chanaged the quotes and could not store the data in file. So anyone please help me.
For more clear, i attach the part of my code
for ($i=0;$i<=10;$i++) { $id=$seqs[$i]; $dbo = new PDO($dbc, $user, $pass); echo $sql = "SELECT * FROM mine_id WHERE locus_id='$id'"; $qry = $dbo->prepare($sql); $qry->execute(); $data = fopen('file.csv', 'w'); while ($row = $qry->fetch(PDO::FETCH_ASSOC)) { fputcsv($data, $row); } } Edited by phpnewbie007, 20 November 2014 - 02:17 AM. Hi there, is it possible to join a variable and a string inside a $_POST variable inside a mysql query (UPDATE in this case) Here is what i am trying to accomplish: $update = "UPDATE mona SET STKFF='$_POST[$counter."NAME"]' WHERE id='$userid'"; if (!mysql_query($update)) { die('Error: ' . mysql_error()); } else echo " <br> Update Complete"; Its the '$_POST[$counter."NAME"]' bit that im worried about, is this possible without having to do: $foo=$counter."NAME" '$_POST[$foo]' Thanks Chris |