PHP - Mysql_fetch_assoc And Echo Problem
I cannot retrieve my records from my table, neither can I display them on my website. I do not even get any errors when I run the script on my site. Any help or tips would be greatly appreciated. My current php script is:
// connection to database server $db_selected = mysql_select_db('database1', $dbc); if (!$db_selected) { die ('Can\'t use database1 : ' . mysql_error()); } $query = "SELECT * FROM chesstable ORDER BY chapter, lastname"; $result = mysql_query($query) or die('Could not connect' . mysql_error()); echo mysql_num_rows($result); If (!$result) { $message = 'Invalid query: ' . mysql_error(); $message = 'Whole query: ' . $query; die ($message); } while($row = mysql_fetch_assoc($result)) { echo $row['firstname'] . " " . $row['lastname'] . $row['chapter']; } mysql_free_result($result); mysql_close($dbc); ?> Similar TutorialsI got this code at the top: Code: [Select] <?php require_once("configuration.php"); $uid = $_GET['uid']; ?> And another main one: Code: [Select] <?php if(isset($_GET['uid'])) { $d = ""; $query = mysql_query("SELECT `username`,`level`,`email,`,`alevel`,`tester` FROM `users` WHERE `id`='$uid'"); if(!mysql_num_rows($query)) echo "No user."; while($d = mysql_fetch_assoc($query)) { ?>   <b>Username:</b> <?php echo $d['username']; ?> <br />   <b>Level:</b> <?php echo $d['level']; ?> <br />   <b>Email:</b> <?php echo $d['email']; ?> <br />   <b>Admin level:</b> <?php echo $d['alevel']; ?> <br />   <b>Tester:</b> <?php echo $d['tester']; ?> <?php } } ?> </font> </div> </center> Now I get this error: Code: [Select] Warning: mysql_num_rows() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\learning\viewuser.php on line 20 No user. Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\learning\viewuser.php on line 22 How can I fix it? Hi all, I have this problem with an echo I have this code, what is working only not all of it, it request a database (ID) TID which is a catorgy. Now I need to TID to be in a place but can't to seem to get it. Here is the code: Code: [Select] <?php defined('_JEXEC') or die('Restricted access'); // no direct access ?> <script type="text/javascript" src="http://maps.google.com/maps/api/js?sensor=false"> </script> <script type="text/javascript"> /* function changeActiveCategory(id,show) { var a_categoryname = "a_category"+id; if(show){ document.getElementById(a_categoryname).className = "active"+id; }else{ document.getElementById(a_categoryname).className = ""; } } */ function changeActiveType(id,show) { var a_typename = "a_type"+id; if(show){ document.getElementById(a_typename).className = "active"+id; }else{ document.getElementById(a_typename).className = ""; } } var markerGroups = { "1": [], "2": [], "3": [], "4": [], "5": []}; function toggleGroup(type) { for (var i = 0; i < markerGroups[type].length; i++) { var marker = markerGroups[type][i]; if (marker.getVisible()) { marker.setVisible(false); changeActiveType(type,false); } else { marker.setVisible(true); changeActiveType(type,true); } } } function changeActiveButton(id) { document.getElementById("idmap").className = "map"; document.getElementById("idsatellite").className = "satellite"; document.getElementById("idterrain").className = "terrain"; var dividname = "id"+id; document.getElementById(dividname).className = id+" "+id+"active"; } var map; var goHome = new google.maps.LatLng(<?php echo $DefaultLat;?>, <?php echo $DefaultLng;?>); function HomeControl(controlDiv, map) { controlDiv.style.padding = '0px 5px'; var controlUI = document.createElement('DIV'); controlUI.style.cursor = 'pointer'; controlUI.setAttribute("class","mapType"); controlDiv.appendChild(controlUI); var controlMap = document.createElement('DIV'); if(map.getMapTypeId()=='roadmap'){ controlMap.setAttribute("class","map mapactive"); }else{ controlMap.setAttribute("class","map"); } controlMap.setAttribute("id","idmap"); controlUI.appendChild(controlMap); var controlsatellite = document.createElement('DIV'); if(map.getMapTypeId()=='satellite'){ controlsatellite.setAttribute("class","satellite satelliteactive"); }else{ controlsatellite.setAttribute("class","satellite"); } controlsatellite.setAttribute("id","idsatellite"); controlUI.appendChild(controlsatellite); var controlterrain = document.createElement('DIV'); if(map.getMapTypeId()=='hybrid'){ controlterrain.setAttribute("class","terrain terrainactive"); }else{ controlterrain.setAttribute("class","terrain"); } controlterrain.setAttribute("id","idterrain"); controlUI.appendChild(controlterrain); google.maps.event.addDomListener(controlUI, 'click', function() { /*map.setCenter(goHome);*/ }); google.maps.event.addDomListener(controlMap, 'click', function() { map.setMapTypeId(google.maps.MapTypeId.ROADMAP); changeActiveButton('map'); }); google.maps.event.addDomListener(controlsatellite, 'click', function() { map.setMapTypeId(google.maps.MapTypeId.SATELLITE); changeActiveButton('satellite'); }); google.maps.event.addDomListener(controlterrain, 'click', function() { map.setMapTypeId(google.maps.MapTypeId.HYBRID); changeActiveButton('terrain'); /*map.setMapTypeId(google.maps.MapTypeId.TERRAIN);*/ }); } function initialize() { var latlng = new google.maps.LatLng(<?php echo $DefaultLat;?>, <?php echo $DefaultLng;?>); var myOptions = { zoom: <?php echo $MapDistance;?>, center: latlng, navigationControl:true, scaleControl:true, mapTypeControl: false, mapTypeId: google.maps.MapTypeId.HYBRID }; var map = new google.maps.Map(document.getElementById("map_canvas"), myOptions); var homeControlDiv = document.createElement('DIV'); var homeControl = new HomeControl(homeControlDiv, map); homeControlDiv.index = 1; map.controls[google.maps.ControlPosition.RIGHT].push(homeControlDiv); var gmarkers = []; var htmls = []; <?php echo 'var ptooltip=new Array();'; echo 'var showhtml=new Array();'; if($items) { $p=0; foreach ($items as $item) { $link = LinkHelper::getLink('properties','showproperty','',$item->CYslug,$item->Sslug,$item->Lslug,$item->Tslug,$item->Pslug); $item->name=str_replace("'","\'",$item->name); echo "\n"."ptooltip[".$p."]=new Array( '".$item->id."','".$item->ref."','".$item->tid."','".$item->name."','".$link."');"."\n"; /*property details*/ if($item->imagename!=NULL){ $img=$item->imagename;}else{$img='noimage.jpg';} $myHtml=''; $myHtml.='<div id="propertydetail"><div class="title"><a href="'. $link.'" title="'.str_replace('"',' ',$item->name).'">'.$item->name.'</a></div><div class="image"><img src="images/properties/images/thumbs/'.$item->id.'/'.$img.'" alt="'.str_replace('"',' ',$item->name).'" width="100" height="75"></div><div class="text">'; if ($item->name_type) { $myHtml.=JText::_($item->name_type).'<br />'; } if ($item->address) { $myHtml.=JText::_($item->address).'<br />'; } $myHtml.='</div>'; $myHtml.='<div class="property_button">'; $myHtml.='<a class="BottomlnkDetail" href="'.$link.'" title="'.$item->id.'">Lees meer...</a>'; $myHtml.='</div>'; $myHtml.='</div>'; echo "\n"."showhtml[".$p."]='".$myHtml."';"."\n"; $myHtml=''; /*end property details*/ $p++; } } ?> var base_Icon=new Array(); base_Icon = '<?php echo JURI::base();?>modules/mod_prop_googlemap/img/beer.png'; <?php if($items) { $x=0; foreach ($items as $item) { ?> var myLatlng = new google.maps.LatLng(<?php echo $item->lat;?>,<?php echo $item->lng;?>); var marker = new google.maps.Marker({ position: myLatlng, map: map, icon: base_Icon }); /*alert (marker.number);*/ marker.setTitle('Klik voor meer info.'); attachSecretMessage(marker, <?php echo $x;?>); markerGroups[HIER MOET TID KOMEN].push(marker); <?php $x++; } } ?> function attachSecretMessage(marker, number) { var infowindow = new google.maps.InfoWindow( { content: showhtml[number], /* disableAutoPan: true,*/ size: new google.maps.Size(50,50) }); google.maps.event.addListener(marker, 'click', function() { infowindow.open(map,marker); /*showInContentWindow(showhtml[number]);*/ }); google.maps.event.addListener(infowindow, 'closeclick', function() { /*infowindow.close(map,marker); showInContentWindow(''); closeInContentWindow();*/ }); /* google.maps.event.addDomListener(controlterrain, 'click', function() { map.setMapTypeId(google.maps.MapTypeId.HYBRID); changeActiveButton('terrain'); }); */ } function showInContentWindow(text) { var sidediv = document.getElementById('content_window'); sidediv.style.visibility = "visible"; sidediv.innerHTML = text; } function closeInContentWindow() { var sidediv = document.getElementById('content_window'); sidediv.style.visibility = "hidden"; } } </script> <div class="prop_googlemap" style="width:960px; height:510px"> <div class="first"> <div class="second"> <div class="third"> <div id="map_canvas" style="width: 950px; height: 500px"></div> <div id="prop_googlemap_categories_all"> <div id="prop_googlemap_categories"> <ul class="categorys"> <li><a id="a_type1" class="active1" href="javascript:void(0)" onclick="toggleGroup('1');">Kroeg</a></li> <li><a id="a_type2" class="active2" href="javascript:void(0)" onclick="toggleGroup('2');">Discotheek</a></li> <li><a id="a_type3" class="active3" href="javascript:void(0)" onclick="toggleGroup('3');">Zaal</a></li> <li><a id="a_type4" class="active4" href="javascript:void(0)" onclick="toggleGroup('4');">test</a></li> </ul> </div> <div style="clear:both"></div> <div class="prop_googlemap_categories_bottom"></div> </div> <div id="content_window"></div> <div id="side_bar"></div> </div> </div> </div> </div> <script type="text/javascript"> initialize(); </script> <div style="clear:both"></div> As you can see he Code: [Select] markerGroups[THIS IS WHERE THE TID SHOULD COME].push(marker);where now it says: "THIS IS WHERE THE TID SHOULD COME" there needs to be a number ( 1 stands for a bar, 2 stands for a disco, 3 for something else. Now this is where I ask for the TID: Code: [Select] echo "\n"."ptooltip[".$p."]=new Array( '".$item->id."','".$item->ref."','".$item->tid."','".$item->name."','".$link."');"."\n"; But it can be me, but can't seem to get it working.... Can't seem to get the right code for "THIS IS WHERE THE TID SHOULD COME" anyone that could help me? Thanks! Hello, I have this syntax on my header div Code: [Select] <p><a href="javascript:ajaxpage('shopcart.php', 'content');" class="difflink"><img src="../stock_photos/shopcart.png"><?php echo '<span class="text">'.writeShoppingCart().'</span>'; ?></a></p> but whatever I try it always output my echo in the next line instead of next to my picture file. How can I do it displayed at the same line? Thank you This php file cant echo in the 2nd php code, can anyone help me about this? 1st php code to connect into the 2nd php code Code: [Select] <?php $Total = $Total + $Amount; } ?> </tr> <tr> <td colspan="3" align="Right">Total</td> <td align ="Center"><?php echo number_format($Total,2);?></td> </tr> 2nd php code Code: [Select] <?php $db = mysql_connect("localhost", "root", ""); mysql_select_db("vinnex",$db); $TransNo = $_POST['TransNo']; $Username = $_POST['Username']; $Date = $_POST['Date']; $Total = $_GET['Total']; echo $Total; $sqltransaction = " INSERT INTO transaction (TransNo, Username, Date, Total) VALUES ('$TransNo', '$Username', '$Date', '$Total')"; $resulttransaction = mysql_query($sqltransaction); ?> Hi im having trouble getting the below code to listen to my css margin settings. Just to clear up the server side of things, can anybody tell me wether or not im using the correct syntax in the code below? Code: [Select] else { echo("<p class=\"passed\">Thankyou for submiting your details, you will be added to our directory shortly</p>"); }} Hi, Im pretty new to php but im have trouble understanding why my script is doing this. Forgive me for silly mistakes! Im creating a login/register website and once the person has logged in they get a message saying "Successful login, welcome username!" The php checks against a database of users in an if statement to check if the password is correct and then prints out the message. But at the end of the message the echo is printing out a 1. I get that because the echo is in an if statement its return true or 1 but I cant get it to not print out the 1. code snippet is - if ($username==$dbusername&&$password==$dbpassword) { echo "Successful login, welcome!"; } Hope that makes sense. Thanks for the help. Neil Having problems with my script echo without form being completed yet. Any help would be appreciated. The echo that is outputting early is "You must enter a Session Name". Code: [Select] <html> <head> <title>ReH-0.1--Create a Session</title> </head> <body bgcolor="575757"> <center> <?php $error = "Could not connect to the database"; mysql_connect('localhost','root','') or die ($error); mysql_select_db('temp') or die ($error); //Set Variables $id = $_POST['id']; $table = $POST['table']; //Check for exsistence if (!$id or !$table) { echo "<font color='#ff0000'>You must enter a Session Name</font>"; } else { echo "Congrats!"; } ?> <h2 style="color:#fff">ReH-0.1 Password Encryption</h2><br><br> <font color="#fff">Before encrypting your password, a session must be started. We need you to enter a personal session name that you will remember so that you password is protected by this session.</font><br><br> <form action="session_start.php" method="POST"> <label for="id" style="color: #fff;">Session Name: <input type="text" maxlength="10" size="10" name="id"></label><br><br> <label for="table" style="color: #fff;">Session Name Confirmation: <input type="text" maxlength="10" size="10" name="table"></label><br><br> <input type="submit" value="Create Session"> </form> </center> </body> </html> ok im building facebook app and im just back to php after months not touched it. this is the code: $image = $facebook->api_client->photos_get(null,null, echo $uid); print_r ($image); echo "<BR>"; echo $image[0]["src"]; the problem is that echo $uid i tried every variation and it does not work! i maybe forget something that i should have done? the $uid containing user id numbers so what i did wrong and if im not wrong i wanted the numbers($uid) Wrapped in quotation marks so i tried: echo ." $uid ". and alot of other please help tnx Is there any way that I can get mysql_fetch_assoc only to run once? Here is my code: Code: [Select] $query = mysql_query("SELECT * FROM items ORDER BY id DESC"); while ($row = mysql_fetch_assoc($query)) { $id = $row['id']; } Code: [Select] $sql = "select * from user_info where us_name='$username' and md5(us_pass)='$userpass'"; $result=mysql_query($sql); $row = mysql_fetch_assoc($result); $count=mysql_num_rows($result); But it is giving error : arning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in C:\...\verify.php on line 19 I am not getting why it is so Not sure what I am doing wrong: Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in /public_html/lq/index.php on line 12 $schedule = "LQ"; include('/public_html/en/header_staff.php'); include("connect.php"); $page = $_SERVER["SERVER_NAME"]; $who = explode(".", $page); $sql = "select * from rp_staff WHERE schedule = $schedule"; $rpstaff = mysql_query($sql); while ($row = mysql_fetch_assoc($rpstaff)) { $alt = $row['dj']; $name = $row['name']; $city = $row['city']; $state = $row['state']; $country = $row['country']; $artists = $row['artists']; $about = $row['about']; $genre = $row['genrev']; $position = $row['position']; } Im having a problem coding for our project . here's the code Code: [Select] <?php $value = $_POST['p']; $host="localhost"; $username="root"; $password=""; $db_name="dbquiz"; mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); $value = stripslashes($value); $value = mysql_real_escape_string($value); $sql='SELECT * FROM `'. $value .'` ORDER BY RAND() LIMIT 100'; $result=mysql_query($sql) or die(mysql_error()); if(result){ while($row = mysql_fetch_assoc($result)) { $q = $row['question']; $c1= "" .$row['choice1']; $c2 ="" .$row['choice2']; $c3 ="" .$row['choice3']; $c4 ="" .$row['choice4']; $a ="".$row['answer']; $questions[] = array($q,$c1,$c2,$c3,$c4,$a); } } include_once("makequiz.php"); ?> AND FOR THE makequiz Code: [Select] <?php if (isset($_POST['sent'])) { for ($i=0;$i<count($questions);$i++) { echo($questions[$i][0]." - "); if ($_POST['q'.$i]=="c") { echo("<b>Correct!</b><br>\n"); $score++; } else { echo("<b>Wrong!</b><br>\n"); } } $percent = number_format(($score/count($questions))*100,2,".",","); echo("<br>".$score." out of ".count($questions)." (".$percent."% right)<br>\n"); } else { echo("<form action=\"#\" method=\"post\">\n"); echo("<input type=\"hidden\" name=\"sent\">\n"); for ($i=0;$i<count($questions);$i++) { echo("<b>".$questions[$i][0]."</b><br><br>\n"); if ($questions[$i][5]==1) { echo("<input type=\"radio\" name=\"q".$i."\" value=\"c\"> ".$questions[$i][1]."<br>\n"); } else { echo("<input type=\"radio\" name=\"q".$i."\" value=\"w\"> ".$questions[$i][1]."<br>\n"); } if ($questions[$i][5]==2) { echo("<input type=\"radio\" name=\"q".$i."\" value=\"c\"> ".$questions[$i][2]."<br>\n"); } else { echo("<input type=\"radio\" name=\"q".$i."\" value=\"w\"> ".$questions[$i][2]."<br>\n"); } if ($questions[$i][5]==3) { echo("<input type=\"radio\" name=\"q".$i."\" value=\"c\"> ".$questions[$i][3]."<br>\n"); } else { echo("<input type=\"radio\" name=\"q".$i."\" value=\"w\"> ".$questions[$i][3]."<br>\n"); } if ($questions[$i][5]==4) { echo("<input type=\"radio\" name=\"q".$i."\" value=\"c\"> ".$questions[$i][4]."<br><br>\n"); } else { echo("<input type=\"radio\" name=\"q".$i."\" value=\"w\"> ".$questions[$i][4]."<br><br>\n"); } } echo("<input type=\"submit\" value=\"Am I Right?!\">"); } ?> When you run the first code.. it's working but when i clicked the submit button this error keeps on showing and i dont know why.. "Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in " can someone help me fix this problem pls? thanks in advance. I am receiving this error: Code: [Select] Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in /home/shannon/public_html/booter/inc/classes/class.db.php on line 17 For the function: Code: [Select] function fetch_array($query) { $array = mysql_fetch_assoc($query); return $array; } Why would I receive that error, if I am just passing a variable? Not a resource or boolean? Thanks. Hi Guys, I keep getting the following error: Parse error: syntax error, unexpected T_ECHO in /Applications/XAMPP/xamppfiles/htdocs/pet/petCatalog1.php on line 51 I have copied the code from a tutorial and cannot for the life of me work out what the problem is with line 51???? I have attached the file and pasted the code below. The real problem line is: echo "<tr><td valign='top' width='15%' style='font-weight: bold; font-size: 1.2em'\n" The full code is attached - btw, inserting the ">" does not help to fix it!! <?php /* Programme: PetCatalog.php * Desc: Lists all pet categories with radio buttons */ ?> <html> <head> <title>Select a Pet Category</title> </head> <body> <?php /* Database login info - Move to an Include */ $host="localhost"; $user="root"; $passwd=""; $dbname="information_schema"; /* Establishes a connection with the databse */ $cxn = mysqli_connect ($host, $user, $passwd, $dbname) or die ("Could not connect to the database JAMIE!"); /* SQL Query to select all categories of pet from the petType column in the pet table */ $query = "SELECT * FROM PetCatalog.petType ORDER BY petType"; $result = mysqli_query($cxn,$query) or die ("Unable to connect to the database JAMIE"); /* Echo of H1 & H2 above the pet table */ echo "<div style='margin-left: .lin'>\n <h1 style='text-align: center'>Pet Catalogue</h1>\n <h2 style='text-align: center'>The following categories of pets are now available</h2>\n"; /* Create form to diplay pet categories */ echo "<form action='ShowPets.php' method='POST'>\n"; echo "<table cellpadding='5' border='1'>"; $counter=1; while ($row = mysqli_fetch_assoc($result)) { extract ($row) echo "<tr><td valign='top' width='15%' style='font-weight: bold; font-size: 1.2em'\n" echo "<input type='radio' name='interest' value='$petType'\n"; if( $counter == 1 ) { echo "checked='checked'"; } echo ">$petType</td>"; echo "<td>$typeDescription</td></tr>"; $counter++; } echo "</table>"; echo "<p>input type='submit' value='Select Pet Type'> </form></p>\n"; ?> </body> </html> Okay, I had a similar problem before.
$users = mysql_query("SELECT * FROM users"); while($users = mysql_fetch_assoc($users)){ echo "<tr><td><p><a href='users.php?id={$users['user_id']}'>{$users['user_name']}</a></p></td><td>{$users['user_clan']}</td><td>{$users['troop_donations']}</td></tr>"; } mysql_fetch_assoc() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\eclipse\test\mysql.php on line 14 this is the error I get if I comment out the if statements. With the If statements I get "query failed" I have to assume the problem is with $extractQuery or $extract ... I just am at a loss for what the problem could be thanks Code: [Select] $connect = mysql_connect("localhost", "root", "") or die(mysql_error()); mysql_select_db("Pickem", $connect) or die(mysql_error()); //extract data from key table $extractQuery = "SELECT * FROM key "; $extract = mysql_query($extractQuery, $connect); if ($extract) { if (mysql_numrows($extract)) { echo $extract; print_r(mysql_fetch_assoc($extract)); } else { echo "no results found"; } } else { echo "query failed"; } I have the following code Code: [Select] $hidden_client_id_query = mysql_query("SELECT id from `pdp_client_info` WHERE lead_id='$post_lead_id'"); $hidden_client_id=mysql_fetch_assoc($hidden_client_id_query); $this->view->hidden_client_id=$hidden_client_id['id']; I get the following error for the above code. I wonder where I am wrong. I am getting the $post_lead_id as expected. Warning: mysql_fetch_assoc() expects parameter 1 to be resource, boolean Would both of the below snippets do the same job, if so why are both ways taught in text books, can one method do somthing that the other cannot? mysql_fetch_array($result, MYSQL_ASSOC)) And... mysql_fetch_assoc($result) Following is the simple code to print name of states. However it doesn't print "Alabama" or in short it misses id=1.Can someone please point out my mistake or tell me why won't it print Alabama and ignoring id = 1 ? While debugging i found out it starts from id= 2. Code - Code: [Select] <?php $sql = "SELECT * FROM state"; $result = mysql_query($sql); ?> <SELECT NAME="state"> <OPTION value="">Select State</option> <?php while($arrayRow = mysql_fetch_assoc($result)) { $strA = $arrayRow["id"]; $strB = $arrayRow["state"]; echo "<option value=\"$strB\">$strB</option>\n"; }?> </SELECT> Sql table image - |