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PHP - Question About Mysql_fetch_array
I do a query and then do a while loop to go threw it, but later down the page I need to loop threw that data again, but nothing is there. Example
while ($row = mysql_fetch_array($sqlmaps)) { //some code } //some code while ($row = mysql_fetch_array($sqlmaps)) { //it dose not work here now } Do need to do that exact same query 2 times to get the data? Similar TutorialsHere is my code Code: [Select] // Show the events for this day: $getEvent_sql = "SELECT cal_title, cal_description, date_format(cal_start_time, '%l:%i %p') as fmt_date FROM calendar WHERE month(cal_start_time) = '".$m."' AND dayofmonth(cal_start_time) = '".$d."' AND year(cal_start_time)= '".$y."' ORDER BY cal_start_time"; $getEvent_res = myload($getEvent_sql); echo "<pre>"; print_r($getEvent_res); echo "</pre>"; if (count($getEvent_res) > 0){ $event_txt = "<ul>"; while($ev = mysql_fetch_array($getEvent_res)){ echo "Inside"; echo $ev; $event_title = stripslashes($ev["cal_title"]); $event_shortdesc = stripslashes($ev["cal_description"]); $fmt_date = $ev["fmt_date"]; $event_txt .= "<li><strong>".$fmt_date."</strong>: ".$event_title."<br/>".$event_shortdesc."</li>"; echo $event_title; } $event_txt .="</ul>"; mysql_free_result($getEvent_res); } else { $event_txt = ""; } if ($event_txt != ""){ echo "<p><strong>Today's Events:</strong></p> $event_txt <hr/>"; } I can't get inside the while loop. The array is populated, I've checked that. The count of $getEvent_res = 3. echo "Inside", and echo $ev give me nothing. Any ideas what I'm doing wrong? $getEvent_res = myload($getEvent_sql); The myload is a custom function that just works, I didn't write it, but I know it works. Hi. I just have a quick question. If I perform a mysql query like the one below that returns all of the rows in a table. How do I then add all of this data into an array. If I use the following code bellow the variable $array will only contain the first row of the table. $rs = mysql_query($query, $conn); $array = mysql_fetch_array ($rs); Hey guys i have a query that select from multiple tables and works fine however i have the same field name in each table called:: member_id but i want to select specifically the member_id from table1 do i need to change the name when i loop through them? i really dont want to do that but if it confuses php unless is there a way when doin a while loop i specify what table i want it from? while($row = mysql_fetch_array($query)){ $member_id = $row['member_id']; } This is a pain. It seems totally logical to me and I had it working before I accidentally saved over the working version (major boo boo). This is two code snippets, first one is how I want it to work - One Query and the loop through results. Second does work, but then its going to have to run mysql 60 times, which could get slow as the site gets more use. $results = mysql_query($query); $response .= "<table><tr>"; if($nit == 0){ for($i = 0; $i < 60; $i++){ $dateCompare = date("Y-m-d", mktime(0,0,0,date("m"),date("d")+$i,date("Y"))); $dateForString = date("D d/m/Y", mktime(0,0,0,date("m"),date("d")+$i,date("Y"))); $response .= "<td height='350px' style='color:white; background-color:grey;'>$dateForString</td>"; while($row = mysql_fetch_array($results)){ $id = $row['id']; $poster = $row['poster']; $date = $row['date']; $response .= "$date : $dateCompare<br>"; if($date == $dateCompare){ $response .= "<td><a href='fullinfo.php?selected_id=$id'><img src='$poster' height='350px' /></a></td>"; } } } } FOLLOWED BY THE SECOND $response .= "<table><tr>"; if($nit == 0){ for($i = 0; $i < 60; $i++){ $dateCompare = date("Y-m-d", mktime(0,0,0,date("m"),date("d")+$i,date("Y"))); $dateForString = date("D d/m/Y", mktime(0,0,0,date("m"),date("d")+$i,date("Y"))); $response .= "<td height='350px' style='color:white; background-color:grey;'>$dateForString</td>"; $results = mysql_query($query); while($row = mysql_fetch_array($results)){ $id = $row['id']; $poster = $row['poster']; $date = $row['date']; if($date == $dateCompare){ $response .= "<td><a href='fullinfo.php?selected_id=$id'><img src='$poster' height='350px' /></a></td>"; } } } } First one results in -it seems - to only go through the while loop once, the first time. So out put is todays date, for the number of results from the database. The second works fine, prints out in order and matches up. I just don't like its inefficiency. Thanks Canty well, i tried to fetch field from my database using the following code: Code: [Select] $result=mysql_query("SELECT * FROM Persons WHERE username='$myusername'" ); while($row = mysql_fetch_row($result)) now that gives me following error: Code: [Select] Warning: mysql_fetch_row() expects parameter 1 to be resource, boolean given in.. Hello if some one be nice to help me out it be most kind... i got this error "Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in/home/23url/public_html/new.php on line 66" the code is - $getbans =mysql_query("SELECT banned FROM $tab[banned];"); $bans = array(); LINE 66 > while($ban=mysql_fetch_array($getbans)) { array_push($bans, $ban[0]); } many thanks Hi I post this thread under php, because I guess that my coding is wrong. I always get the following message Quote Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/content/g/w/o/gwolff2005/html/admin/update.php on line 23 Update data in mysql The code is Code: [Select] <!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"> <?php $host="xxxx"; // Host name $username="xxxx"; // Mysql username $password="xxxxx; // Mysql password $db_name="xxxxx"; // Database name $tbl_name="sp_users"; // Table name // Connect to server and select database. mysql_connect("$host", "$username", "$password")or die("cannot connect"); mysql_select_db("$db_name")or die("cannot select DB"); // get value of id that sent from address bar $id=$_GET['user_id']; // Retrieve data from database $sql="SELECT * FROM '$sp_users' WHERE id='$user_id'"; $result=mysql_query($sql); $result=mysql_query("select * from '$sp_users' WHERE id='$user_id'"); $rows = mysql_fetch_array($result); ?> <table width="400" border="0" cellspacing="1" cellpadding="0"> <tr> <form name="form1" method="post" action="update_ac.php"> <td> <table width="100%" border="0" cellspacing="1" cellpadding="0"> <tr> <td> </td> <td colspan="3"><div align="center" class="style1">Update data in mysql </div></td> </tr> <tr> <td align="center"> </td> <td align="center"> </td> <td align="center"> </td> <td align="center"> </td> </tr> <tr> <td align="center"> </td> <td align="center"><span class="style7">Name</span></td> <td align="center"><span class="style7">Lastname</span></td> <td align="center"><span class="style7">Email</span></td> </tr> <tr> <td> </td> <td align="center"><input name="name" type="text" id="name" value="<? echo $rows['user_first_name']; ?>"></td> <td align="center"><input name="lastname" type="text" id="lastname" value="<? echo $rows['user_surname']; ?>" size="15"></td> <td><input name="email" type="text" id="email" value="<? echo $rows['user_login']; ?>" size="15"></td> </tr> <tr> <td> </td> <td><input name="id" type="hidden" id="id" value="<? echo $rows['user_id']; ?>"></td> <td align="center"><input type="submit" name="Submit" value="Submit"></td> <td> </td> </tr> </table> </td> </form> </tr> </table> <? // close connection mysql_close(); ?> </body> </html>What am I doing wrong. Thanks for your help! please tell me the reason that why mysql_fetch_array does not work inside a function for example the following works well in its current form $sql="SELECT * FROM $tbl_13"; $query=mysql_query($sql); while($rs=mysql_fetch_array($query)){ $p0 = $rs["pfid"]; $pm = $rs["pm"]; } but when i put it inside a function it gives me an error at function defone() { $sql="SELECT * FROM $tbl_13"; $query=mysql_query($sql); while($rs=mysql_fetch_array($query)){ $p0 = $rs["pfid"]; $pm = $rs["pm"]; } return $pm; } the error message is... Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in... Please identify the problem.. i'm an experience programmer in ruby and perl but new in php I am having a problem with my script not returning anything! I echo the variable usually it will say ARRAY and it has something in it.. Nothing at all it coming up no errors I have put or die in it, nothing. I tried error_reporting(E_ALL); , still nothing. If someone can help me with this, I will greatly appreciate it. ~AJ error_reporting(E_ALL); //Get Catagorys and echo $getCatagorys = mysql_query("SELECT * FROM `catagory`") or die("Query failed with error: ".mysql_error()); while($row = mysql_fetch_array($getCatagorys, MYSQL_ASSOC)){ echo ' <tr> <td class="catagoryTD" width="550">'.$row['catagoryTitle'].'</td> <td class="catagoryTD" width="75"><center>Threads</center></td> <td class="catagoryTD" width="75"><center>Posts</center></td> <td class="catagoryTD" width="170"><center>Last Post</center></td> </tr>'; //Get SubCatagorys and echo $catagoryID = $row['catagoryID']; $getSubCatagory = mysql_query("SELECT * FROM `subcatagory` where catagoryID = '$catagoryID'"); while($subrow = mysql_fetch_array($getSubCatagory, MYSQL_ASSOC)){ $subCatagory = $subrow['subcatagoryID']; //Get Stats for number of threads $threads = mysql_query("SELECT * FROM entry WHERE subcatagoryID = '$subCatagory'") or die(mysql_error()); $num_threads = mysql_num_rows($threads); //Get Stats for number of entrys(posts) $posts = mysql_query("SELECT * FROM posts WHERE subcatagoryID = '$subCatagory'") or die(mysql_error()); $num_posts = mysql_num_rows($posts); //Find newest post and find what thread it is from $lastPost = mysql_query("SELECT * FROM `posts` ORDER BY `posts`.`postCreated` DESC") or die(mysql_error()); while($lastPostReturn = mysql_fetch_array($getSubCatagory, MYSQL_ASSOC)){ echo $lastPostReturn; } //Grab User and thread number and date $lastPostUser = $lastPostReturn['postUser']; $lastPostCreated = $lastPostReturn['postCreated']; $lastPostEntry = $lastPostReturn['entryID']; //Find the title of the thread and generate URL\ $lastPostThread = mysql_query("SELECT * FROM `entry` WHERE entryID = '$lastPostEntry'"); while($lastPostThreadReturn = mysql_fetch_array($lastPostThread, MYSQL_ASSOC)){ } echo ' <tr> <td><a href="catagory.php?id='.$subrow['subcatagoryID'].'">'.$subrow['subcatagoryTitle'].'</a></td> <td><center>'.$num_threads.'</center></td> <td><center>'.$num_posts.'</center></td> <td>'.$lastPostThreadReturn['text'].'<br>'.$lastPostCreated.'<br>'.$lastPostUser.'</td> </tr>'; } } Hi everyone, Bit of a random question and im sure there is good reason so if someone could in lighten me that would be great ! With regards to the below code, it seems to just create an infinite loop when you put mysql_fetch_array. Just want to know what wrong with doing this? and why it act in this way ? Code: [Select] $sql = mysql_query("SELECT * FROM cities"); $results = mysql_fetch_array($sql); while($row = $results) { echo $row['City']."<br />"; } thanks Guys Hi guys! I inputted text into the field 'TSuperQuestion' of my table. Now I am trying to echo that text. I get the following error: Quote Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /webusers/ph6theo/public_html/Code-sandbox.php This is my code: Code: [Select] <?PHP $con = mysql_connect("localhost","ph6theo", "xxxx"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("ph6theo_testDB") or die(mysql_error()); $result = mysql_query("SELECT * FROM Q&ATable"); $row = mysql_fetch_array($result); echo $row['TSuperQuestion']; ?> Can anyone help? I have some OOP code im working on and I have some interactions with the database. However, mysql_fetch_array fails to be working. Have I missed something? Code: [Select] class core { var $array_batsmen = ''; var $array_bowlers = ''; // Query to get batsmen function get_batsmen() { return $this->array_batsmen = mysql_query('SELECT id, name, ability, strength FROM wtw_players WHERE team_id = 1 ORDER BY id ASC'); } // Query to get bowlers function get_bowlers() { $this->array_bowlers = mysql_query('SELECT id, name, ability, strength FROM wtw_players WHERE team_id = 2'); } // Generate array of batsmen to be used function generateBatsmen() { while ($array = mysql_fetch_array($this->array_batsmen)) { echo $array['name']; } } } The functions are called later normally, i've only included the necessary parts. Basically I have the var $array_batsmen which stores the mysql_query. This is then later used with a mysql_fetch_array to return names from the earlier query. But nothing happens! It doesn't return anything (don't worry, I have a return later on). The loop just doesn't seem to work :/ Hey guys, i'm sure this is simple i have a script that always brings back the same error and I beleive it's going to be the php version im on the code is Code: [Select] function check_user( $u ) { $res = query("SELECT COUNT(*) FROM `usertable` WHERE `userid`=$u"); list($total_rows) = mysql_fetch_array($res) or die (mysql_error()); if( $total_rows > 0 ) return true; else return false; } the error is Code: [Select] Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/bofs/public_html/thief/inc/userinc.php on line 10 You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1 any idea's Hi All, I am sorry to bother you but wondered if you might be able to spot my mistake... When I click on this page http://www.freepspwallpaper.co.uk/free-psp-wallpapers/free-babes-psp-wallpapers/ if returns an error as below (look carefully its hidden behind the header color). Im a bit of a newbie and cannot seem to remove it. Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/freepspw/public_html/dbconnect.php on line 22 Can anyone help? Peter Hey, I have been having a problem with this code for a while now where I connect to the MySQL database to retrieve the desired information, but always get an error message. I have tried to correct it myself by changing bits of the code and searched the internet, but still have no idea how to fix it. Any help would be appreciated. The error is as follows: Notice: An error occured1: in <filelocation> on line 18 Warning: mysql_fetch_array() expects parameter 1 to be resource, null given in <filelocation> on line 21 Notice: An error occured2: in <filelocation> on line 15 My code as is follows: require_once('../mysqli_connect.php') ; $query = "SELECT * FROM database WHERE mem_id='".($_GET['id'])."'"; $result = @mysql_query($dbc, $query) OR trigger_error("An error occured1: " . mysql_error()); $row = mysql_fetch_array($result) OR trigger_error("An error occured2: " . mysql_error()); echo "<b>Date: </b> ".$row['date']; echo "<br><b>ID: </b> ".$row['mem_id']; Thanks. Hi, I am a little confused on the difference the mysql_fetch_array and mysql_query? Whats the difference between those two? Thanks Hey guys, Im trying to post data on a page from multiple rows. Here is how it looks: <?php //FIRST ROW $result0 = mysql_query("SELECT * FROM test_mysql")or die(mysql_error()); $row0 = mysql_fetch_array($result0); //SECOND ROW $result1 = mysql_query("SELECT * FROM test_mysql LIMIT 1,1")or die(mysql_error()); $row1 = mysql_fetch_array($result1); //THIRD ROW $result2 = mysql_query("SELECT * FROM test_mysql LIMIT 2,2")or die(mysql_error()); $row2 = mysql_fetch_array($result2); ?> <?php echo $row0['name']; echo "<br>"; echo $row0['lastname']; echo "<br>"; echo $row0['email']; echo "<br>"; ?> ----- <?php echo "<br>"; echo $row1['name']; echo "<br>"; echo $row1['lastname']; echo "<br>"; echo $row1['email']; echo "<br>"; ?> ----- <?php echo "<br>"; echo $row2['name']; echo "<br>"; echo $row2['lastname']; echo "<br>"; echo $row2['email']; echo "<br>"; ?> As you can see, I am needing to make a new MYSQL_QUERY all the time because I have to change the limit on it so It shows data on another row. Is there any way to do this easier by putting the row I want to show inside the $row2['email']. I have about 6 rows and I need to show data on a page. This is how it looks. Can anyone show me a simple way of doing this? I want to use all five results separately in order to use in a Jquery slider. Code: [Select] <?php $a = "SELECT * FROM blog_posts ORDER BY id DESC LIMIT 5"; $query = mysql_query($a) or die (mysql_error()); $query_results = mysql_fetch_array($query); ?> I know if i was going to grab vars from an normal array i would just use... Code: [Select] <?php echo $array_reults['0'] ?> Can any one help me with this problem? I'm trying to pull out a list of id's into a simple array, but I seem to be getting is a multidimensional array. All I need is a simple array with the list of id's. What am I doing wrong? //Get list of all sentence id's selected for test in database $allSentenceQuery = mysql_query("SELECT id FROM associations WHERE validate='checked'") or die("First Query: ".mysql_error()); while ($allSentenceResult = mysql_fetch_array($allSentenceQuery)){ $allSentences[] = $allSentenceResult['id']; //echo "allSentences: ".$allSentenceResult['id']; } And what I get: "Array ( => Array ( [1] => 2 [2] => 3 etc...." Thanks for your help! |
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