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My Result Having Dot...how To Get The Values...	View Content

$name = $name . "<option value=". $row['name'] . ">" .$row['name'] . "</option>";

but the name is "Mr. Robert",it does have dot symbol...how to overcome? please help

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Trying To Post Updated Values To Db And Getting "fatal Error: Uncaught Typeerror: Mysqli_fetch_assoc(): Argument #1 ($result) Must Be Of Type Mysqli_result, Bool Given"

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I have a table with a list of users and an edit button and delete button. When the edit button is pressed on a site it passes the user_id as p_id to the page that catches it and displays the data. The problem is when I click on the "update user" button, I get the following error:

Warning: Undefined variable $the_user_id in C:\xampp\htdocs\3-19-21(2) - SafetySite\admin\edit_user.php on line 10

Fatal error: Uncaught TypeError: mysqli_fetch_assoc(): Argument #1 ($result) must be of type mysqli_result, bool given in C:\xampp\htdocs\3-19-21(2) - SafetySite\admin\edit_user.php:14 Stack trace: #0 C:\xampp\htdocs\3-19-21(2) - SafetySite\admin\edit_user.php(14): mysqli_fetch_assoc(false) #1 {main} thrown in C:\xampp\htdocs\3-19-21(2) - SafetySite\admin\edit_user.php on line 14

The weird thing is I had another update user page with a table I created that ran the query to update the table in the database just fine. But as I created it, it didn't look all that great so I recreated the page and used a bootstrap table because of the much cleaner look. Both pages have the exact same PHP code, the only difference is the bootstrap table I added in. So I'm really at a loss with this. Other than the table and PHP code, there is a script at the bottom of the page for the table itself to allow for searching within the table, i'll include that as well. The PHP code is as follows:

<?php

//THE "p_id" IS BROUGHT OVER FROM THE EDIT BUTTON ON VIEW_ALL_USERS

if (isset($_GET['p_id'])) {

$the_user_id = $_GET['p_id'];

}

// QUERY TO PULL THE SITE INFORMATION FROM THE p_id THAT WAS PULLED OVER

$query = "SELECT * FROM users WHERE user_id = $the_user_id ";

$select_user = mysqli_query($connection,$query);

//SET VALUES FROM ARRAY TO VARIABLES

while($row = mysqli_fetch_assoc($select_user)) {

$user_id = $row['user_id'];

$user_firstname = $row['user_firstname'];

$user_lastname = $row['user_lastname'];

$username = $row['username'];

$user_email = $row['user_email'];

$user_phone = $row['user_phone'];

//$user_image = $row['user_image'];

$user_title_id = $row['user_title_id'];

$user_role_id = $row['user_role_id'];

}

THE UPDATE QUERY CODE....................................................................................................................

<?php

if(isset($_POST['update_user'])) {

$user_id = $_POST['user_id'];

$user_firstname = $_POST['user_firstname'];

$user_lastname = $_POST['user_lastname'];

$username = $_POST['username'];

$user_email = $_POST['user_email'];

$user_phone = $_POST['user_phone'];

//$user_image = $_POST['user_image'];

$user_title_id = $_POST['user_title_id'];

$user_role_id = $_POST['user_role_id'];

$query = "UPDATE users SET ";

$query .= "user_id = '{$user_id}', ";

$query .= "user_firstname = '{$user_firstname}', ";

$query .= "user_lastname = '{$user_lastname}', ";

$query .= "username = '{$username}', ";

$query .= "user_email = '{$user_email}', ";

$query .= "user_phone = '{$user_phone}', ";

//$query .= "user_image = '{$user_image}', ";

$query .= "user_title_id = '{$user_title_id}', ";

$query .= "user_role_id = '{$user_role_id}' ";

$query .= "WHERE user_id = '{$the_user_id}' ";

$update_user = mysqli_query($connection,$query);

if(! $update_user) {

die("QUERY FAILED" . mysqli_error($connection));

}

THE "UPDATE USER" BUTTON THE USER CLICKS ON TO UPDATE....................................................................................................................

<button class="btn btn-primary" type="submit" name="update_user">Update User</button>

</div>

Any Help is Greatly Appreciated!

Edited March 23 by Zserene

Getting Date Result, But Not Expecting Any. Why Does It Give Result?

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I am really lost here with this date issue of mine. The below code is the last part of a query:
Code: [Select]
$defendercheck_3 = $row_checkifattacked3['atdate'];
$defendercheck1_3 = strtotime("$defendercheck_3");
$defendercheck2_3 = date("D", $defendercheck1_3);
The query does not return any results as expected, but when echoing the various steps I get following:

echo "$defendercheck3"; = nothing (as expected)

echo "$defendercheck1_3"; = nothing (as expected)

echo "$defendercheck2_3"; = result! (NOT expected)

why does it return anything on "date("D", $defendercheck1_3)" when "$defendercheck1_3" is blank?

How To Insert Two Input Values And Multi Data Textarea Values In One Query?

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Dear All Members

here is my table data..

(4 Columns/1row in mysql table)

id order_no order_date miles
310 001 02-15-2020 MI,108.53
Oh,194.57
PA,182.22
WA,238.57

How to split(miles) single column into (state, miles) two columns and output like following 5 columns /4rows in mysql using php code.

(5 Columns in mysql table)
------------------------------------------------------------------------------------

id order_no order_date state miles

310 001 02-15-2020 MI 108.53

310 001 02-15-2020 Oh 194.57

310 001 02-15-2020 PA 182.22

310 001 02-15-2020 WA 238.57

------------------my php code -----------

<?php
//invoice.php
include('../includes/config.php');

if(isset($_POST["add"]))
{
$order_no=$_GET['order_no'];
$order_date=$_GET['order_date'];
$miles=$_POST['miles'];

$miles = explode("\r\n", $_POST["miles"]);
$newdata = "'" . implode("','", $_POST["miles"]) . "'";
$sql ="INSERT INTO tble_states_miles(order_no,order_date, miles) VALUES(:order_no, :order_date, :newdata)";

$query = $dbh->prepare($sql);
$query-> bindParam(':order_no', $order_no, PDO::PARAM_STR);
$query-> bindParam(':order_date', $order_date, PDO::PARAM_STR);
$query-> bindParam(':miles', $newdata, PDO::PARAM_STR);

$lastInsertId = $dbh->lastInsertId();
if($lastInsertId)
{
echo "<script type='text/javascript'>alert('Added Sucessfull!');</script>";
echo "<script type='text/javascript'> document.location = 'ifta_miles_calculationlist_added.php'; </script>";
}
else
{
$error="Something went wrong. Please try again";
}

if($query->execute()) {

$sql = "update tis_invoice set flag='1' where order_no=:order_no";
$query = $dbh->prepare($sql);
$query-> bindParam(':order_no', $order_no, PDO::PARAM_STR);

$query->execute();

}
}
?>

----------------- my form code ------------------

<?php
               $order_no=$_GET['order_no'];
       $sql = "SELECT * from tis_invoice where order_no = :order_no";
       $query = $dbh -> prepare($sql);
       $query->execute(
array(
':order_no' => $_GET["order_no"]
)
);
       $result = $query->fetchAll();
       foreach($result as $row)
?>

<form method="post">
<table align="center" style="width: 55%">
                       <tr><td>Invoice Number</td></tr>
                       <tr><td><input type="text" name="order_no" value="<?php echo $row["order_no"]; ?>" readonly=""></td></tr>
                       <tr><td>Invoice Date</td></tr>
                       <tr><td><input type="text" name="order_no" value="<?php echo $row["order_date"]; ?>" readonly=""></td></tr>
                       <tr><td>
<textarea name="miles" class="form-control" rows="10" placeholder="" required></textarea>
</td></tr>
                       <tr><td>

<input type="submit" name="add" class="btn btn-primary" value="Add" />
</td></tr></table>
</form>

-- Can any one help how to correct my code..present nothing inserted on table

Thank You

Edited February 8, 2020 by karthicbabu

Table Issue - Multiple Location Values For User Pushes Values Out Of Row Instead Of Wrapping In Cell

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Hi,

My company has 240+ locations and as such some users (general managers) cover multiple sites. When I run a query to pull user information, when the user has multiple sites to his or her name, its adds the second / third sites to the next columns, rather than wrapping it inside the same table cell. It also works the opposite way, if a piece of data is missing in the database and is blank, its pull the following columns in. Both cases mess up the table and formatting. I'm extremely new to any kind of programming and maybe this isn't the forum for this question but figured I'd give it a chance since I'm stuck.

The HTML/PHP code is below:

<thead>

<tr>

<th>FirstName</th>

<th>LastName</th>

<th>Username</th>

<th>Phone #</th>

<th>Location</th>

<th>Title</th>

<th>Actions</th>

</tr>

</thead>

<tbody>

<?php

//QUERY TO SELECT ALL USERS FROM DATABASE

$query = "SELECT * FROM users";

$select_users = mysqli_query($connection,$query);

// SET VARIABLE TO ARRAY FROM QUERY

while($row = mysqli_fetch_assoc($select_users)) {

$user_id = $row['user_id'];

$user_firstname = $row['user_firstname'];

$user_lastname = $row['user_lastname'];

$username = $row['username'];

$user_phone = $row['user_phone'];

$user_image = $row['user_image'];

$user_title_id = $row['user_title_id'];

$user_role_id = $row['user_role_id'];

// POPULATES DATA INTO THE TABLE

echo "<tr>";

echo "<td>{$user_id}</td>";

echo "<td>{$user_firstname}</td>";

echo "<td>{$user_lastname}</td>";

echo "<td>{$username}</td>";

echo "<td>{$user_phone}</td>";

//PULL SITE STATUS BASED ON SITE STATUS ID

$query = "SELECT * FROM sites WHERE site_manager_id = {$user_id} ";

$select_site = mysqli_query($connection, $query);

while($row = mysqli_fetch_assoc($select_site)) {

$site_name = $row['site_name'];

echo "<td>{$site_name}</td>";

}

echo "<td>{$user_title_id}</td>";

echo "<td>{$user_role_id}</td>";

echo "<td class='table-action'>

</td>";

//echo "<td><a href='users.php?source=edit_user&p_id={$user_id}'>Edit</a></td>";

echo "</tr>";

}

<tr>

<td>FirstName</td>

<td>LastName</td>

<td>Username</td>

<td>Phone #</td>

<td>Location</td>

<td>Title</td>

</td>

</tr>

</tbody>

<tfoot>

<tr>

<th>FirstName</th>

<th>LastName</th>

<th>Username</th>

<th>Phone #</th>

<th>Location</th>

<th>Title</th>

</tr>

</tfoot>

</table>

Searching Multidimensional For Values, Then Outputting Other Values In Array

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Mysql Values Clause, Variable Values From File

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Hi all,

Just curious why this works:

Code: [Select]
while (($data = fgetcsv($handle, 1000, ",")) !== FALSE){
$import="INSERT into $prodtblname ($csvheaders1) values('$data[0]','$data[1]','$data[2]','$data[3]','$data[4]','$data[5]','$data[6]')";
}
And this does not:

$headdata_1 = "'$data[0]','$data[1]','$data[2]','$data[3]','$data[4]','$data[5]','$data[6]'";
while (($data = fgetcsv($handle, 1000, ",")) !== FALSE){
$import="INSERT into $prodtblname ($csvheaders1) values($headdata_1)";
}it puts $data[#'s] in the database fields instead of the actual data that '$data[0]','$data[1]'... relates to.

I wrote a script to create the values in $headdata_1 based on the number of headers in $csvheaders1 but can't seem to get it working in the sql statement.

Thanks

No Result

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Basically I want this to do do is:

if (there is no result found in the query){
echo "nothing for this brand";
}else{
echo this link;
}

this is my code...im just confused on what i want the if statement to to ask for.

any help would be greatly appreciated

Code: [Select]
<?php
$mypro = mysql_query("SELECT * FROM Sheet1 WHERE pro_catagory LIKE '%People%' ORDER BY pro_id DESC");
while($row=mysql_fetch_array($mypro)) {
if(Not sure what to put here){
echo "There are no listings for the brand right now";
}else{
print "- <a href='proList.php?proNumber=".$row['pro_id']."'>".$row['pro_name']."</a><br />";
}
}
?>

Sql If Result ='s

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I'm writing an admin panel for my site (to manage users, etc)

I selected the user-rights from the DB, but how can I check it the user-rights is equal to 3, and if it is allow access?

Why Am I Not Getting A Result?

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at the moment i have my page:

addevent.php

i have written:
Code: [Select]
$q = "SELECT username "
."FROM ".TBL_USERS." ORDER BY ASC";
$result = $database->query($q);

on the database.php page here is what query looks like:
Quote

/**
* query - Performs the given query on the database and
* returns the result, which may be false, true or a
* resource identifier.
*/
function query($query){
return mysql_query($query, $this->connection);
}

then i want to display $result in a drop down menu so i wrote back on addevent.php:

Code: [Select]
<select name="Select1">
<option value="<? echo("$result"); ?>"><? echo("$result"); ?></option>
</select>

Why Am I Not Getting A Result???

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Ok i call a function like so ...
<?php PunchTime($monday,"in","1",$etime_who); ?>

Below is the function, if you look closely i echo the query after it's been ran, and that query works 100% since i copied & pasted the query into the database and it returns a result that's needed. But for some bloody reason it doesn't work via code.

function PunchTime($pdate,$type,$num,$ewho)
{
if($type == "in")
{
if($num == "1")
{
$etime_col = "PUNCH_IN";
}
elseif($num == "2")
{
$etime_col = "PUNCH_IN2";
}
}
elseif($type == "out")
{
if($num == "1")
{
$etime_col = "PUNCH_OUT";
}
elseif($num == "2")
{
$etime_col = "PUNCH_OUT2";
}
}

include_once('oracleCON.php');

$etime_qry = "SELECT $etime_col FROM tablename WHERE who= '$ewho' AND punch_date = '$pdate'";
echo $etime_qry;
$etime_go = oci_parse($conn, $etime_qry);
oci_execute($etime_go);
$etime_res = oci_fetch_assoc($etime_go);

$time = $etime_res[$etime_col];
if($time == "") $time = "--";

echo $time;
}

All i get are "--" even though some should return a time stamp.

Always Same Result

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i have a game script and it's always bringing back the same result and i cant see why..

all i get when the script it ran is

[/code]echo '<div class=noticeBox><h2>Sorry! No available member found![/code]

the code is below and can provide the functions and db if needed, but cant see what's wrong.

Code: [Select]
<?php

include '../header.php';

echo '<div id=AjaxSubDiv>';
echo '<div id="actionDiv">';

include '../statusbar.php';

echo '<div id=contentBox>';

// Show Fight Result
if( !empty($_GET[action]) AND $_GET[action]=='attack' ) {
echo $str;

// Fight Lists
echo '<P class=pageheader>Fight</P>';
echo '<P>Its time for action! Fight with rival thieves and grab their cash. You will require 1 strength for each fight and you will gain some cash and exp if you win the fight. However you will just earn some exp if you loose!! Each fight will decrease your health so keep an eye on health level because if your health goes below zero you will die and loose everything...</P>';

$res = query("SELECT COUNT(*) FROM `usertable` WHERE `userid`!=$user");
list($total) = mysql_fetch_array($res);
if( $total>0 ) {

$res = query("SELECT * FROM `usertable` WHERE `userid`!=$user LIMIT 0,20");

$viewplayer = Player::getById($row[userid]);

echo '<tr>';
echo '<td width=120px>';
echo '<P class=titleP><a href="#" onclick="load_menu_val(\'AjaxSubDiv\', \'profile\', '.$row[userid].');return false;"><fb:profile-pic uid="'.$row[userid].'" size="square" linked="false" width="75px" class="userImg" /></a></P>';
echo '<P class=titleP><a href="#" onclick="load_menu_val(\'AjaxSubDiv\', \'profile\', '.$row[userid].');return false;"><fb:name uid="'.$row[userid].'" linked="false" /></a></P>';
echo '</td>';

echo '<td width=120px>';
echo '<P><B>Level:</B> '.$viewplayer->level.', '.get_level_title($player->level).'</P>';
echo '</td>';

echo '<td width=120px>';
echo '<P><B>Crew:</B> '.$viewplayer->crew.'</P>';
echo '</td>';

echo '<td width=300px>';
if( $viewplayer->health > 20 ) {
echo '<form id="FightForm'.$row[userid].'" method="POST" class="itemForm">';
echo '<input type=hidden name=user value="'.$user.'">';
echo '<input type=hidden name=opid value="'.$row[userid].'">';
echo '<input type=hidden name=action value="dofight">';
echo '<P><input type=button onclick="do_submit(\'FightForm'.$row[userid].'\',\'http://www.bofs.us/thief/thiefmaster/fight/post_fight.php\', \'actionDiv\'); return false;" value="Fight Now" class="submitButton" /></P>';
echo '</form>';
}
else {
echo '<P class=titleP>Too Weak To Fight</P>';
}
echo '</td>';
echo '</tr>';
}
echo '</table>';
echo '</center>';
}
else {
echo '<div class=noticeBox><h2>Sorry! No available member found!</h2></div>';
print"$res";

}

echo '</div>'; // end of contentBox

echo '</div>'; // end of actiondiv
echo '</div>'; // end of AjaxSubDiv

?>

Only Returning First Result

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hello, anybody able to help me with why this is only returning the first staff member's hours?

Code: [Select]
<?php

if(isset($_POST['view']))
{

$y3=$_POST['y'];
$m3=$_POST['m'];
$d3=$_POST['d'];

$pdate=$y3."-".$m3. "-".$d3;
$pdate1 = date( 'D M j', strtotime($pdate) );
} else {

$pdate =  date('Y-m-d', strtotime("-1 day") );
$pdate1 = date( 'D M j', strtotime($pdate) );

}
echo "<table border='1' style='border-collapse: collapse' bordercolorlight='#000000' bordercolordark='#000000' width='98%' align='center'>";
echo "<tr><td width='100%' colspan='9' align='center'><b>Timesheets For $pdate1</b></td></tr>";

$result = mysql_query("SELECT * FROM staff ORDER BY name");
while($row = mysql_fetch_array($result))
{
$eid=$row['eid'];
$name=$row['name'];

  echo "<tr>";
  echo "<td align='left' colspan='9'><b>" . $name . "</b></td>";
  echo "</tr>";

echo "<tr>
<th align='center'>Date</th>
<th align='center'>Job Number</th>
<th align='center' width='30%'>Service Report</th>
<th align='center'>Sign In Time</th>
<th align='center'>Sign Out Time</th>
<th align='center'>Lunch</th>
<th align='center'>Time Billed</th>
<th align='center'>Estimated</th>

</tr>";

$result3 = mysql_query("SELECT * FROM timesheet WHERE date = '$pdate' AND eid = '$eid'");
$talltime = 0;
while($row3 = mysql_fetch_array($result3))
{

  $tid=$row3['id'];
  $tdate=$row3['date'];
  $tjobnumber=$row3['jobnumber'];
  $teid=$row3['eid'];
  $tdescription=$row3['description'];
  $tsignin=$row3['start'];
  $tfinish=$row3['finish'];
  $tlunch=$row3['lunch'];
  $tkms=$row3['kms'];
  $tschednum=$row3['schednum'];


  $tdate1 = date( 'M j, Y', strtotime($tdate) );

    $tsignin1 = date( 'g:i a', strtotime($tsignin) );
$tfinish1 = date( 'g:i a', strtotime($tfinish) );

if( empty($tfinish) ) {
$tfinish2="<i>In Progress";
} else {
$tfinish2="$tfinish1";
}

$shortid=substr($tjobnumber, 5, -1); //remove the first 5 characters and minus the last character

$result = mysql_query("SELECT * FROM jobs WHERE id = '$shortid'");
while($row = mysql_fetch_array($result))
{
$jstatus=$row['status'];

}

$result4 = mysql_query("SELECT * FROM schedule WHERE id = '$tschednum'");
while($row4 = mysql_fetch_array($result4))
{

  $sid=$row4['id'];
  $sdate=$row4['date'];
  $seid=$row4['eid'];
  $sjobnumber=$row4['jobnumber'];
  $sstarttime=$row4['starttime'];
  $sstatus=$row4['status'];
  $setime=$row4['etime'];

}

$log_in_time_string = strtotime($tsignin);
$log_out_time_string = strtotime($tfinish);
$difference_in_seconds = ($log_out_time_string - $log_in_time_string);
$tbilled = ($difference_in_seconds / 3600);


  if($tbilled < 0)
{
    $tbilled1 = $tbilled + 24;
} else {
    $tbilled1=$tbilled;
}

  $tbilled2 = number_format(round($tbilled*4)/4,2);
  $tbilled3 = $tbilled2 - $tlunch ;
  $talltime += $tbilled3;


  echo "<tr>";
  echo "<td align='center'>" . $tdate1 . "</td>";
  echo "<td align='center'>" . $tjobnumber . "</td>";
  echo "<td align='center'>" . $tdescription . "</td>";
  echo "<td align='center'>" . $tsignin1 . "</td>";
echo "<td align='center'>" . $tfinish2 . "</td>";
echo "<td align='center'>" . $tlunch . " hour</td>";
echo "<td align='center'>";
     if ($tbilled3 > $setime ) {
echo "<font color='red'><b>*** " . $tbilled3 . " hours ***</b></font>";
} else {
echo "" . $tbilled3 . " hours";
}
echo "</td>";
echo "<td align='center'>" . $setime . " hours</td>";

  echo "</tr>";

}
   echo "<tr>";

  echo "<td colspan='6' align='right'><b>Totals :</td>";
  echo "<td align='center' colspan='2'>";
     if ($talltime > "8" ) {
echo "<font color='red'><b>*** " . $talltime . " hours ***</b></font>";
} elseif ($talltime < "0" ) {
echo "<font color='red'><b>Not Signed Out</b></font>";
} else {
echo "" . $talltime . " hours";
}
echo "</td>";

  echo "</tr>";

   echo "<tr>";
echo "<td align='center' colspan='8' bgcolor='#D9FFD9'><hr></td>";
  echo "</tr>";

}

echo "</table>";

include 'close.php';

?>

Help With Order A Result

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Hey everybody, i'm building a website for the softball league i play in.... so far the website is pretty good... with all the stats and profiles.... i'm having a little problem with the ORDER of the leaders in the categories i have to calculate the result...

for example... avg. is the result of the total hits divided by the total number of at bats... both numbers are coming from the database... the problem comes when i have to order the leaders from first to last...

this is my php:

Code: [Select]
$sql = "select profile_id, SUM(total_hits) AS hitstotal, SUM(ab) AS abtotal from batters WHERE year = '2010' GROUP BY profile_id";

$cons = mysql_query($sql);
while ($result = mysql_fetch_array($cons)) {
$sumaab = $resultad['abtotal'];
$sumahits = $resultad['hitstotal'];
$p_id = $resultad['profile_id'];
$prom = round(($sumahits."") / ($sumaab.""),3);

that gives me all the batting averages for the players.

and here is my question:
how can i order that $prom result from first to last?

Display Just One Result

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i have two tables, and i want to display data from the most recent entry of the two(i.e. display just one entry). i'm joining together 2 tables so that i can get this entry.Both tables contain the date, so i'm ordering the results of the query so that the most recent comes first.
My tables a
TOPIC
topic_id
topic _name
section_sub_id
date

POST
post_id
post_name
section_sub_id
date

I'm using the following code to get the last result, but nothing is being echoed
$join =mysql_query("SELECT topic.*, post.* FROM topic LEFT JOIN post ON topic.section_sub_id=post.section_sub_id WHERE post.section_sub_id = " .$row2['section_sub_id']. " ORDER BY topic.date") or die("Select Error :" . mysql_error());
$latest= mysql_fetch_row($join);
echo $latest['post.date'];
i'm not sure if there is something wrong with the sql, or with the php. ANy help would be great

Null Result

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hi all,

i have a script in where the user enters there post code and it will provide a price for delivery, how would i echo "please contact us for price" if the db record for the price is (null) or empty??

<?php
mysql_connect ("localhost", "root","password")  or die (mysql_error());
mysql_select_db ("postcode");
$term = $_POST['term'];
$sql = mysql_query("SELECT * FROM uk_postcodes, zones WHERE postcode LIKE '%$term%' AND zone_id = zone_large ");
while ($row = mysql_fetch_array($sql)){
    echo '<br/> Postcode: '.$row['postcode'];
    echo '<br/> Town: '.$row['town'];
    echo '<br/> County: '.$row['county'];
    echo '<br/> Small: £'.$row['price_s'];
    echo '<br/> Medium: £'.$row['price_m'];
    echo '<br/> Large: £'.$row['price_l'];
    echo '<br/><br/>';
    }

?>

Mysql Result = ''?

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How can I check if a returned mysql value is equal to '' i.e. nothing?

I keep getting an error where the page won't load because the returned value is '' so i need to check for it

Formatting One-to-many Sql Result

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Hi there,

So I have data which I am fetching from two tables link in a one-to-many relationship using the following code:

sql = "select authcourses.id as id, title, date_format(closingdate, '%d.%m.%y') as date, name from authcourses inner join authorities on authorityid = authorities.id order by authorityid";
$result = mysqli_query($link, $sql);
if(!$result)
{
  $error = 'Unable to get list of authorities';
  include '../../error.html.php';
  exit();
}

while ($row = mysqli_fetch_array($result))
{
  $authcourses[] = array('name' => $row['name'], 'title' => $row ['title'], 'date' => $row['date'], 'id' => $row['id']);
}

I then output the data in a table like this:
Code: [Select]
<table>
<thead><th>Title</th><th>Closing Date</th><th>Action</th></thead>
<tbody>
<?php foreach ($authcourses as $authcourse): ?>
<tr>
<form action="?" method="post">
<tr>
<td><?php htmlout($authcourse['name']); ?></td> 
<td><?php htmlout($authcourse['title']); ?></td>
<td><?php htmlout($authcourse['date']); ?></td>
<td>
<input type="hidden" name="id" value="<?php echo $authcourse['id']; ?>" />
<input type="submit" name="action" value="Edit" />
<input type="submit" name="action" value="Delete" />
</td>
</tr>
</form>
<?php endforeach; ?>
</tbody>
</table>

The htmlout() is a custom function which is basically "echo htmlspecialchars($str)", and you can ignore the form stuff.

I get an out put like this:
York, Course 1, 2011-02-15
York, Course 2, 2011-03-01
Manchester, Course 3, 2011-06-17
Manchester, Course 4, 2011-08-12
Derby, Course 5, 2011-01-10
Barnet, Course 6, 2011-08-19
Barnet, Course 7, 2011-06-23

etc etc...

What I want is something like this:
York:
Course 1, date
Course 2, date

Manchester:
Course 3, date
Course 4, date

Derby:
Course 5, date

Barnet:
Course 6, date
Course 7, date

etc..

I guess I'd like to load an array for each of the authorities (cities) with the course info, then I can output it in my form like this:
Code: [Select]
<?php foreach ($authcourses as $authcourse): ?>
<h1><?php htmlout($authcourse['name'] ?>: </h1>
<?php foreach ($courses as $course): ?>
<p><?php htmlout($course['title'];?><p>
<p><?php htmlout($course['date'];?></p>
<?php endforeach; ?>
<?php endforeach; ?>

I hope that makes sense... Any help would be great.

Cheers,
Mike

One Result Been Inserted Twice

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Basically the following code works fine, exepct when it comes to the last result it inserts it twice, example:

(3,17),(4,17),(5,17),(5,17)

Any clues as to why? Thanks

Code: [Select]
if(count($addIDs_ary) > 0)
{
$str = "";
foreach($addIDs_ary as $val)
{
$str .= "({$val},{$playerID}),";
if(end($addIDs_ary) == $val)
{
$str .= "({$val},{$playerID})";
}
}
echo $str; // (val,val), (val,val), (val,val) etc..
$query = "INSERT INTO hitlist
(hit_id,player_id) values $str";

How Can I Get This Result Of The Query?

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Hello all. I am using a premade login script which has some very powerful function built in, the one I am looking at using is as follows:

  /** 
    * query - Performs the given query on the database and 
    * returns the result, which may be false, true or a 
    * resource identifier. 
    */ 
   function query($query){ 
      return mysqli_query( $this->connection, $query); 
   }

The issue is, displaying the results. I call is via the following code:

$database->query('SELECT LastName, Firstname FROM Patrons WHERE LastName = "Delude"');

But I have no idea how to display it. I have attempted to echo it out, but I get the following error in the logs: PHP Catchable fatal error: Object of class mysqli_result could not be converted to string

Result Paging

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Hello, everyone. I am having some trouble figuring this out. I am trying to create a paging bar(so to speak) to display results by page number.
Trying to get the paging bar to go as follows...
Requested Page The resulting paging bar
Page 1 1 2 3 4 5
Page 2 1 2 3 4 5
Page 3 1 2 3 4 5
Page 4 1 2 3 4 5
Page 5 4 5 6 7 8
Page 6 4 5 6 7 8
Page 7 4 5 6 7 8
Page 8 7 8 9 10 11
......
This is what I have right now that I am trying to get to work. I have changed this so many times and been working on it for 2 days. I think I'm just confusing my self. This can't be that hard..
Code: [Select]
<?php
echo '<html>';
//---these values changed based on query----
$Page=$_GET["page"]; //the page number requested
$AvailablePages=5; //number of pages available after the page requested

//---calculations---
$StartPage=round($Page/5)*5;
$EndPage=ceil($StartPage/5)*5 ; //the last page number to be displayed in paging
if($StartPage<5) {$StartPage=1;}

//just looking at variables
echo 'page is: ' . $Page . '<br>';
echo '<br>Starting @: ' . $StartPage;
echo '<br>Ending @: ' . $EndPage . '<br>';

//--create the paging bar--
$PagingBar='';
while($StartPage<=$EndPage)
{
if($Page==$StartPage) {
$PagingBar=$PagingBar . $StartPage . '';
}
else {
$PagingBar=$PagingBar . '
<a href="test.php?page=' . $StartPage . '">' . $StartPage . '</a>
';
}
$StartPage++;
}

//---show the paging bar--
echo $PagingBar;
echo '</html>';
?>