mysql_fetch_array not fetching complete data? - php

I have a code which fetches data from a mysql table and converts it into pdf document, the code is working fine except it is skipping row 1.
Here is the code from which i have removed the pdf generation process since the problem is in the loop which is fetching data.
Please help.
<?php
session_start();
if(isset($_SESSION['user']))
{
$cr = $_POST['cour'];
$s = $_POST['sem'];
require('fpdf.php');
include('../includes/connection.php');
$sql = "SELECT * FROM `student` WHERE AppliedCourse ='$cr'";
$rs = mysql_query($sql) or die($sql. "<br/>".mysql_error());
if(!mysql_fetch_array($rs))
{
$_SESSION['db_error'] = "<h2><font color = 'RED'>No such course found! Pease select again.</font></h2>";
header('Location: prinrepo.php');
}
else {
for($i = 0;$i <= $row = mysql_fetch_array($rs);$i++)
{
$formno[$i] = $row ['FormNo'];
$rno[$i] = $row ['rollno'];
$snm[$i] = $row ['StudentNm'];
$fnm[$i] = $row ['FathersNm'];
$mnm[$i] = $row ['MothersNm'];
$addr[$i] = $row['Address'];
$pic[$i] = $row['imagenm'];
$comm[$i] = $row['SocialCat'];
echo $formno[$i]."<br />";
echo $rno[$i]."<br />";
echo $snm[$i]."<br />";
echo $fnm[$i]."<br />";
echo $mnm[$i]."<br />";
echo $addr[$i]."<br />";
echo $pic[$i]."<br />";
echo $comm[$i]."<br />";
echo "<br />";
}
}
mysql_close($con);
}
?>

You are fetching the first row outside of your for() loop then you miss it.
After mysql_query() your should use mysql_num_rows() to check if there are any rows in your result and then fetch them in the for loop.
More info here : http://php.net/manual/fr/function.mysql-num-rows.php
Your code would look like this :
$sql = "SELECT * FROM `student` WHERE AppliedCourse ='$cr'";
$rs = mysql_query($sql) or die($sql. "<br/>".mysql_error());
if(0 == mysql_num_rows($rs)) {
$_SESSION['db_error'] = "<h2><font color = 'RED'>No such course found! Pease select again.</font></h2>";
header('Location: prinrepo.php');
} else {
for($i = 0;$i <= $row = mysql_fetch_array($rs);$i++)
{
// Your code
}
}

Related

possible ?: mysql row to an if condition

hi guys im trying to insert a mysql data to a variable that will set an if condition depending on the result. is this possible, am i doing it right? what is the right way to do it ? what i want to achieve is to validate if there's a equal value given by the user inside my mysql rows.
$db = mysql_connect('localhost','test','');
if (!$db)
{
print "<h1>Unable to Connect to MySQL</h1>";
}
$dbname = 'test';
$btest = mysql_select_db($dbname);
if (!$btest)
{
print "<h1>Unable to Select the Database</h1>";
}
$sql_statement = "SELECT * ";
$sql_statement .= "FROM registered_email ";
$result = mysql_query($sql_statement);
$outputDisplay = "";
$myrowcount = 0;
if (!$result) {
$outputDisplay .= "<br /><font color=red>MySQL No: ".mysql_errno();
$outputDisplay .= "<br />MySQL Error: ".mysql_error();
$outputDisplay .= "<br />SQL Statement: ".$sql_statement;
$outputDisplay .= "<br />MySQL Affected Rows: ".mysql_affected_rows()."</font><br />";
}
else{
$numresults = mysql_num_rows($result);
for ($i = 0; $i < $numresults; $i++)
{
$row = mysql_fetch_array($result);
$id = $row['id'];
$sentEmailClients = $row['email'];
$outputDisplay.= "".$sentEmailClients."<br />";
}
}
and here what im trying to achieve, btw is $clientEmail has a default values so dont worry about that.
if($clientEmail === $outputDisplay){
...... some codes..........
}
else{
....... some codes.......
}
you can use mysql row to compare with your user input. you can add condition, while you'r getting row value for the email inside the loop.
$email_exist = 0;//define the default value.
for ($i = 0; $i < $numresults; $i++)
{
$row = mysql_fetch_array($result);
$id = $row['id'];
$sentEmailClients = $row['email'];
$outputDisplay.= "".$sentEmailClients."<br />";
//my code start here
if($sentEmailClients == $clientEmail)
$email_exist = 1;
}
//outside the loop
if($email_exist == 1) {
//..........write some code.......
}else{
//........write some code.......
}
why don't you use a while loop?
make sure to update to mysqli_* because mysql_* is deprecated and is going to get removed on php 7.0
$email_exist = 0;//define the default value.
while ( $row = mysql_fetch_assoc($result) ) // you are using associative array and not the indexed once tho you should go for mysql_fetch_assoc
{
$id = $row['id'];
$sentEmailClients = $row['email'];
$outputDisplay.= "".$sentEmailClients."<br />";
//my code start here
if($sentEmailClients == $clientEmail)
$email_exist += 1; //maybe it exist more than once?
}
//outside the loop
if($email_exist == 1) {
//..........write some code.......
}else{
//........write some code.......
}
or you can do something more simple like this
$query = "select email from tablename where email='$clientemail'";
$result = mysql_query($query);
$count = mysql_num_rows($result);
if($count > 0) {
// email exists
} else {
// doesn't exist
}

Displaying data horizontally using php and mysql

Hi I'm attempting to display data retrieved from a mysql table horizontally in an html table using php. The code below works well except for the fact that it leaves out the first record (starts at the second record) in my database. I'm sure it has something to do with the counter but I can't seem to figure out how to get it to stop doing this. If anyone can point out my error I'd really appreciate it!
$items = 5;
$query = "SELECT * FROM members ";
$result = mysql_query($query)
or die(mysql_error());
$row = mysql_fetch_array($result);
if (mysql_num_rows($result) > 0) {
echo '<table border="1">';
$i = 0;
while($row = mysql_fetch_array($result)){
$first_name = $row['first_name'];
if ($i==0) {
echo "<tr>\n";
}
echo "\t<td align=\center\">$first_name</td>\n";
$i++;
if ($i == $items) {
echo "</tr>\n";
$i = 0;
}
}//end while loop
if ($i > 0) {
for (;$i < $items; $i++) {
echo "<td> </td>\n";
}
echo '</tr>';
}//end ($i>0) if
echo '</table>';
}else {
echo 'no records found';
}
try and remove the 1st
$row = mysql_fetch_array($result);
you are calling it twice, that's why it skips 1 row in your while loop
try this simpler.
$items = 5;
$query = "SELECT * FROM members ";
$result = mysql_query($query) or die(mysql_error());
if (mysql_num_rows($result) > 0) {
echo '<table border="1">';
while($row = mysql_fetch_array($result)){
$first_name = $row['first_name'];
echo "<tr>";
for ($i=0 ; $i <= $items ;$i++) {
echo "<td align='center'>".$first_name."</td>";
}
}//end while loop
echo "</tr>";
echo '</table>';
}else{ echo 'no records found'; }
I have run into this issue before. Try the do while loop instead. Example
do {
// code
} while($row = mysql_fetch_array($result)); //end while loop
$row = mysql_fetch_array($result);
1) remove this line of code from ur scripts
2) only use while loop code instead.

PHP - Not echoing data from a MySQL database, but no errors?

So I have this PHP code:
Note: I do use mysqli_connect() further up.
$result = mysqli_query($con,"SELECT * FROM `smf_messages` WHERE `id_board` = 18");
if(!$result) {
echo "<center><p>Couldn't fetch news posts. Error code 2.</p></center>";
mysqli_close($con);
} else {
$posts = array();
$topicbdy = array();
while($row = mysqli_fetch_array($result,MYSQLI_ASSOC))
{
$posts[$row['id_topic']] = $row['id_topic'];
$topicbdy[$row['id_msg']] = $row['id_msg'];
}
$display = max($posts);
$display2 = min($topicbdy);
$qry = "SELECT * FROM `smf_messages` WHERE `id_board` = 18 AND `id_topic` = " . $display . " AND `id_msg` = " . $display2;
$result2 = mysqli_query($con,$qry);
//echo $qry;
if(!$result2) {
echo "<center><p>Couldn't fetch news posts. Error code 3.</p></center>";
} else {
while($show = mysqli_fetch_array($result,MYSQLI_ASSOC))
{
echo "<center><h1>" . $show['subject'] . "</h1></center><br /><br />";
echo "<center>" . $show['body'] . "</center><br />";
}
}
mysqli_free_result($result);
mysqli_free_result($result2);
mysqli_close($con);
It's supposed to get the latest topic out of the database for my SMF-based forum from the news board, by getting the highest topic id, but the lowest post id. It seems to be doing the query just fine, as I don't get any errors, but it doesn't show the subject or body. What should I do?
Your $result variable is wrong for second query fetch. For your second query
while($show = mysqli_fetch_array($result,MYSQLI_ASSOC))
Should be
while($show = mysqli_fetch_array($result2,MYSQLI_ASSOC))
^

how do I traverse rows in a table

I have a table named members with 3 rows. The following code attempts to display all the rows in the members table. It displays the 1'st record 3 times instead of displaying each record once.
<?php
$con = new mysqli("localhost", "root", "jce123", "profile");
if (mysqli_connect_errno()) {
printf("Connection failed: %s\n", mysqli_connect_error());
exit();
}
$sql = "SELECT * FROM members";
$res = mysqli_query($con,$sql);
$num = mysqli_num_rows($res);
$array = mysqli_fetch_assoc($res);
$keysarray = array_keys($array);
$valuearray = array_values($array);
for ($i=0; $i < $num; $i++) {
for($j = 0; $j < count($array); $j++) {
echo "<table>";
echo "<tr><td>".$keysarray[$j].": </td><td>".$valuearray[$j]."</td></tr>";
echo "</table>";
}
echo "<br><br>";
}
mysqli_close($con);
?>
I've updated this per suggestions:
$sql = "SELECT * FROM members";
$res = mysqli_query($con,$sql);
$num = mysqli_num_rows($res);
$array = mysqli_fetch_assoc($res);
while ($row = mysqli_fetch_assoc($res)) {
foreach($array as $key => $value){
echo "<table>";
echo "<tr><td>".$key.": </td><td>".$value."</td></tr>";
echo "</table>";
}
echo "<br><br>";
}
That's because, your $num = 3, count($array) = 3 and the outer for loop has no bearing on inner for loop.
Where you have $array = mysqli_fetch_assoc($res);, you will only receive one row from your database. You need something like:
while ($row=mysqli_fetch_assoc($res))
{
// processing for each row
}
Any basic tutorial will tell you this. Even the PHP docs provide an example.
you can use do something like this..
while ($row=mysqli_fetch_assoc($res))
{
// processing for each row
e.g
echo $row['id'];
echo $row['name'];
etc
}
try this... this peace of code is not tested but just to give you an idea...
$sql = "SELECT * FROM members";
$res = mysqli_query($con,$sql);
$num = mysqli_num_rows($res);
while ($row = mysqli_fetch_assoc($res)) {
echo "<table>";
echo '<tr><td> id = "'.$row['id'].'":
</td><td> name = "'.$row['name'].'"</td></tr>";
echo "</table>";
}

Update Multiple Rows (PHP + MySQL)

I am working on a lead management system - and as the database for it grows the need for more bulk functions appears - and unfortunately I am getting stuck with one of them. The database stores many different leads - with each lead being assigned to a specific closer; thus the database stores for each lead the lead id, name, closer name, and other info. The main lead list shows a checkbox next to each lead which submits the lead id into an array:
<input type=\"checkbox\" name=\"multipleassign[]\" value=\"$id\" />
Now this all goes to the following page:
<?php
include_once"config.php";
$id = $_POST['multipleassign'];
$id_sql = implode(",", $id);
$list = "'". implode("', '", $id) ."'";
$query = "SELECT * FROM promises WHERE id IN ($list) ";
$result = mysql_query($query);
$num = mysql_num_rows ($result);
if ($num > 0 ) {
$i=0;
while ($i < $num) {
$closer = mysql_result($result,$i,"business_name");
$businessname = mysql_result($result,$i,"closer");
echo "$closer - $businessname";
echo"<br>";
++$i; } } else { echo "The database is empty"; };
echo "<select name=\"closer\" id=\"closer\">";
$query2 = "SELECT * FROM members ";
$result2 = mysql_query($query2);
$num2 = mysql_num_rows ($result2);
if ($num2 > 0 ) {
$i2=0;
while ($i2 < $num2) {
$username = mysql_result($result2,$i2,"username");
$fullname = mysql_result($result2,$i2,"name");
echo "<option value=\"$fullname\">$fullname</option>";
++$i2; } } else { echo "The database is empty"; }
echo "</select>";
?>
I want to be able to use the form on this page to select a closer from the database - and then assign that closer to each of the leads that have been selected. Here is where I have no idea how to continue.
Actually - i got it. I don't know why I didn't think of it sooner. First off I passed the original $list variable over to the new page - and then:
<?php
include_once"config.php";
$ids = $_POST['list'];
$closer = $_POST['closer'];
$query = "UPDATE `promises` SET `closer` = '$closer' WHERE id IN ($ids) ";
mysql_query($query) or die ('Error updating closers' . mysql_error());
echo "A new closer ($closer) was assigned to the following accounts:";
$query = "SELECT * FROM promises WHERE id IN ($list) ";
$result = mysql_query($query);
$num = mysql_num_rows ($result);
if ($num > 0 ) {
$i=0;
while ($i < $num) {
$businessname = mysql_result($result,$i,"business_name");
echo "<li>$businessname";
++$i; } } else { echo "The database is empty"; };
?>
The updated page before this:
$query = "SELECT * FROM promises WHERE id IN ($list) ";
$result = mysql_query($query);
$num = mysql_num_rows ($result);
if ($num > 0 ) {
$i=0;
while ($i < $num) {
$closer = mysql_result($result,$i,"business_name");
$businessname = mysql_result($result,$i,"closer");
echo "$closer - $businessname";
echo"<br>";
++$i; } } else { echo "The database is empty"; };
echo "<form name=\"form1\" method=\"post\" action=\"multiple_assign2.php\">";
echo "<input type=\"hidden\" name=\"list\" value=\"$list\" />";
echo "<select name=\"closer\" id=\"closer\">";
$query2 = "SELECT * FROM members ";
$result2 = mysql_query($query2);
$num2 = mysql_num_rows ($result2);
if ($num2 > 0 ) {
$i2=0;
while ($i2 < $num2) {
$username = mysql_result($result2,$i2,"username");
$fullname = mysql_result($result2,$i2,"name");
echo "<option value=\"$fullname\">$fullname</option>";
++$i2; } } else { echo "The database is empty"; }
echo "</select>";
echo "<input name=\"submit\" type=\"submit\" id=\"submit\" value=\"Reassign Selected Leads\">";
?>
After you select the leads and submit the form , your script should show them in a list with hidden inputs (with name=leads[] and value=the_lead's_id) and next to each lead there will be a dropdown box () which will be populated with all the closers.
After choosing and sending the second form your script will "run" all-over the leads' ids array and update each and every one of them.
Got the idea or you want some code?

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