PHP - Excluding Ids From Select

When I run this query to exclude records where id = $userarray['id'],

$q = "SELECT * FROM users WHERE inst_id = '". $userarray['inst_id'] ."' AND id != '". $userarray['id'] ."' ";


it actually ONLY returns records with $userarray['id']  as opposed to excluding them.  Isn't the way to exclude them to use != ?

Hi Everyone,
I have a rather stupid question, but I haven't been able to find the correct method / syntax to do this. I want to allow users to upload a maximum of 5 photos - less then 5 is OK too. I've been testing the following script and haven't gotten it to ignore error # 4 (no file uploaded):

Code: [Select]
   if ($_FILES['userfile']['error'] == 1 || 2 || 3 || 6 || 7)   {
 
    echo 'Problem: ';
    switch ($_FILES['userfile']['error'])   {

      case 1: echo 'File exceeded upload_max_filesize';
  break;
      case 2: echo 'File exceeded max_file_size';
  break;
      case 3: echo 'File only partially uploaded';
  break;
  case 6:   echo 'Cannot upload file: No temp directory specified.';
  break;
  case 7:   echo 'Upload failed: Cannot write to disk.';
  break;
    }
    exit;
  }
I will be using a while loop and incrementing through the script (this is simpler version). Does anyone know the proper syntax for the IF statement ' if ($_FILES['userfile']['error'] == 1 || 2 || 3 || 6 || 7)" ?
Thanks much,
Rick

Hello, I need to hide products with same title from WooCommerce shop page and i'm trying to achieve that with a custom loop.

For example i have 5 products called "Plex" and the only diferrence between them is the SKU field and the color. I don't care which of the same product will be displayed in the shop page, i just want to display only one of all these products.

I created a loop which is working and does hide products with same name but i have a problem. If i set posts_per_page=8 it shows less because it counts the hidden products also.

$args = array(
   'post_type'         => 'product',
   'posts_per_page'    => 8
);
$query = new WP_Query($args);
$list = array();

while ($query->have_posts()) : $query->the_post(); 
if(in_array(get_the_title(), $list)){ continue; }
   $list[] = get_the_title();
?>

<li><?php wc_get_template_part( 'content', 'product' ); ?></li>

<?php endwhile;
wp_reset_postdata();

P.S. I don't want to remove these products, i just want to hide them from shop loop!

Any ideas what is the problem with the loop? Is there a better way to achieve that?

I must be brain dead...I've searched everywhere and can't seem to find a snippet for this. I'm just looking for some code to show a directory listing, but only certain files (by extension) and I don't want to show any sub-directories.

I thought I might have been able to do this in the .htaccess file, but I can't find anything that will get rid of the sub-directories.

So, I figure there must be a way to get a file listing, check the extension, add it to the page, or skip it if it's a sub-directory or doesn't match the extension.

Or, I could just be totally, way off...anyway, any help would be greatly appreciated...I'm fairly new to php.

hirealimo.com.au/code1.php

this works as i want it:
Quote

SELECT * FROM price INNER JOIN vehicle USING (vehicleID) WHERE vehicle.passengers >= 1 AND price.townID = 1 AND price.eventID = 1



but apparelty selecting * is not a good thing????

but if I do this:
Quote

SELECT priceID, price FROM price INNER JOIN vehicle....etc


it works but i lose the info from the vehicle table.

but how do i make this work:
Quote

SELECT priceID, price, type, description, passengers FROM price INNER JOIN vehicle....etc


so that i am specifiying which colums from which tables to query??

thanks

I have 2 queries that I want to join together to make one row  

Dear All,

I wish to have 2 drop down boxes, Country Select Box and Locality Select Box. The locality select box will be affected by the value chosen in the country select box.

All is working fine except that the locality select box is not being populated.
I know that the problem is in the sql statement

WHERE country_id='$co'

because i am having an error that $co is an undefined variable.

All the rest works fine because i have replaced the $co variable directly with a number (say 98) for a particular country id and it worked fine. In what way can i define this variable $co so that it is accepted by my sql statement?

Thank you for your help in advance.

MySQL Tables indicated below:

CREATE TABLE countries(
country_id INT(3) UNSIGNED NOT NULL AUTO_INCREMENT,
country_name VARCHAR(30) NOT NULL,
PRIMARY KEY(country_id),
UNIQUE KEY(country_name),
INDEX(country_id),
INDEX(country_name))
ENGINE=MyISAM;

CREATE TABLE localities(
locality_id INT(10) UNSIGNED NOT NULL AUTO_INCREMENT,
country_id INT(3) UNSIGNED NOT NULL,
locality_name VARCHAR(50),
PRIMARY KEY (locality_id),
INDEX (country_id),
INDEX (locality_name))
ENGINE=MyISAM;

Extract PHP script included below:

// connect to database
require_once(MYSQL);

if(isset($_POST['submitted']))

{

// trim the incoming data

/* this line runs every element in $_POST through the trim() function, and assigns the returned result to the new $trimmed array */
$trimmed=array_map('trim',$_POST);

// clean the data

$co=mysqli_real_escape_string($dbc,$trimmed['country']);

$lc=mysqli_real_escape_string($dbc,$trimmed['locality']);

}

?>

<form action="form.php" method="post">

<p>Country
<select name="country">
<option>Select Country</option>

<?php

$q="SELECT country_id, country_name
FROM countries";

$r=mysqli_query($dbc,$q) or
trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

while($row=mysqli_fetch_array($r))

{

$country_id=$row[0];
$country_name=$row[1];
echo '<option value="' . $country_id . '"';
if(isset($trimmed['country']) && ($trimmed['country']==$country_id))
echo 'selected="selected"';
echo '>' . $country_name . '</option>\n';

}

?>

</select>
</p>

<p>Locality
<select name="locality">
<option>Select Locality</option>

<?php

$ql="SELECT locality_id, country_id, locality_name
FROM localities
WHERE country_id='$co'
ORDER BY locality_name";

$rl=mysqli_query($dbc,$ql) or
trigger_error("Query: $q\n<br />MySQL Error: " . mysqli_error($dbc));

while($row=mysqli_fetch_array($rl))

{

$locality_id=$row[0];
$country_id=$row[1];
$locality_name=$row[2];
echo '<option value="' . $locality_id . '"';
if(isset($trimmed['locality']) && ($trimmed['locality']==$locality_id))
echo 'selected="selected"';
echo '>' . $locality_name . '</option>\n';

}

// close database connection
mysqli_close($dbc);

?>

</select>
</p>

<p><input type="submit" name="submit" value="Submit" /></p>

<input type="hidden" name="submitted" value="TRUE" />

</form>

Hi,

I'm trying to make a 3 page registration form.

On the 3rd page I want it to echo back all the options they have selected.  I can get the input type to work but having problems with the selects.

First Page
Code: [Select]
<form action="choose.password.php" method="post">
eMail: <input type="text" name="eMail" /><br />
Forename: <input type="text" name="fname" /><br />
Surname: <input type="text" name="lname" /><br />
Date of Birth: <input type="text" name="dob" /><br />
Gender: <select name="gender">
<option value="male">Male</option>
<option value="female">Female</option>
</select>
<br />
I am not a lawyer: <input type="checkbox" name="lawyer" value="lawyer">
<br />
<br />
<input type="submit" value="Next" />

 
</form>

Second Page
Code: [Select]
<form action="confirm.details.php" method="post">
<p>email address: <?php echo $_POST["eMail"]; ?></p>
<p>password: <input type="password" name="eMail" /></p>
<p>Confirm Password: <input type="password" name="eMail" /></p>
<p>Hint: <input type="text" name="hint" /></p>
<p>My email is correct, I am happy to be sent occassional emails, and i am not a family lawyer. <input type="checkbox" name="lawyer" value="lawyer"></p>
<input type="hidden" name="fname" value="<?php echo ($_POST["fname"]); ?>" />
<input type="hidden" name="eMail" value="<?php echo ($_POST["eMail"]); ?>" />
<input type="hidden" name="lname" value="<?php echo ($_POST["lname"]); ?>" />
<input type="hidden" name="dob" value="<?php echo ($_POST["dob"]); ?>" />
<select style="display:none;" name="gender" <?php echo ($_POST["gender"]); ?> />
<p><input type="submit" value="Next" /></p>

Third page
Code: [Select]
<p>Please confirm your details are correct to register.</p>
<p>Email Address: <?php echo $_POST["eMail"]; ?>
<p>Forename: <?php echo $_POST["fname"]; ?></p>
<p>Surname: <?php echo $_POST["lname"]; ?></p>
<p>Date of Birth: <?php echo $_POST["dob"]; ?></p>
<p>Gender: <?php echo $_POST["gender"]; ?></p>

How can I make the select work and echo it back on the third page?

Thanks

hi. just started a website talkietaco.com and at the momment my code selects the first row and displays it. this is all well and good but when i add a new row to the table it gos to the bottom. I want to be able to select the last row and echo it out. Any ideas? would i need to add an id row or something?
Heres the code and thank you in advance for any help people can offer.

Code: [Select]
<?php


$con = mysql_connect("localhost","","");



if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
  
mysql_select_db("mainbase", $con);


$result = mysql_query("SELECT * FROM matteroffact");

echo "<table border='0'>
<tr>
<th></th>
</tr>";

$row = mysql_fetch_array($result) or die(mysql_error());
echo "<tr>";
echo "<td><strong>" . $row['question']. "</strong>  ". $row['answer']; "</td>";
echo "<tr>";

echo "</table>";


?>

Code: [Select]
id player_id nat nt_caps
13740 28664 97 24
13741 28664 68 0
13742 28664 79 0
16252 42904 15 40
16253 42904 68 0
16254 42904 241 0

That's how my table looks. I want to select the player_id's that have either nt_caps = "0" for every nat OR player_id's that have nt_caps != "0" only for nat = "68".

The SQL query I try to use is:

SELECT player_id FROM x WHERE nat = '68' AND (nat != '68' AND nt_caps = '0')

But then I get player_id '42904' and '28664' because they both have 1 entry that matches the query but I don't want them because they have nt_caps for another nat than nat "68".

I hope you understand what I try to achieve.

Hello dear friends,

If i've database table has named as info (id, name, number)
and i wanna say select from info where name must has value not empty

Code: [Select]
select * from info where name='xxxxwhatxxx'
i've tired $name and $1 and 1 but all not working how to say where name must has value and not empty


example if i've
Code: [Select]
(id, name, number) values (1, 'manal', '4')
(id, name, number) values (2, 'shimaa', '7')
(id, name, number) values (3, '', '3')
and i wanna say sellect from info where name has value so it shows me only data of id "1" and "2" only

thank you so much
it would helps me alot to improve myself

Hi, what I want to figure out is for instance a person has registered with their country e.g. England. Now if I echo the country in a select box giving the person an option to change their country and Showing the person which currently they have selected already. The select box shows two England`s to select from. Could some one tell me how can I have one of each country and echo their already selected country from the database. I don't know how to explain any better what I am after but just basically there are two Englands showing one which is already selected (echoing from the mysql database)  and one is already in the select box.  Any help is much appreciated thank you.

ok here's my problem $Sql1 returns two values: 8 and 10 and these numbers get put into a <select>. So far so good. I assign a onchange to it. When i select 8 it makes the changes but when I select 10 nothing happends. I preciate some help.

Code: [Select]
<?php
require("status.php");
require("id.php");

$Link = mysql_connect($Host, $User, $Password);
mysql_select_db('sportsportal', $Link);

$Sql1 = "select distinct week, (select max(week) from coupons where user='$User') max from coupons where user='$User'";
$Result1 = mysql_query($Sql1, $Link);

print "<tr>";
print "<td align=left valign=top>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td>";
print "<td align=left valign=top>";

print "<select name=current_week onchange=window.location='coupon.php?curwk='+this.value>";
while($Row1 = mysql_fetch_array($Result1)){

//if($Row1[week] == $Row1[max]){
//print "<option value='$Row1[week]' selected>Vecka $Row1[week]</option>";
//} else {
//print "<option value='$Row1[week]'>Vecka $Row1[week]</option>";
//}

print "<option value='$Row1[week]'>Vecka $Row1[week]</option>";

}
print "</select>";

print "<p></td>";
print "</tr>";

if(isset($_REQUEST['curwk'])){
$Curwk = $_REQUEST['curwk'];
} else {
$Curwk = 0;
}

$Sql = "select home, away, home_score, away_score, winner from coupons where user='$User' and week='$Curwk'";
$Result = mysql_query($Sql, $Link) or die(mysql_error());

while($Row = mysql_fetch_array($Result)){

if(@$Row[home] == @$Row[winner]){
print "<tr>";
print "<td align=left valign=top>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td>";
print "<td align=left valign=top><b>$Row[home]</b> - $Row[away] $Row[home_score]-$Row[away_score]</td>";
print "</tr>";
} else {
print "<tr>";
print "<td align=left valign=top>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</td>";
print "<td align=left valign=top>$Row[home] - <b>$Row[away]</b> $Row[home_score]-$Row[away_score]</td>";
print "</tr>";
}

}

mysql_close($Link);
?>

Hi Guys   I have a table which in it's shortened form has the following columns:   id  |  postID  |  title  |  content  |  version   The column for postID has a number that can be shared by multiple rows - differentiated by version number.   I want to run a query to select all records that are like a given keyword (i.e. %LIKE%) but where results share the same postID I only want to return the highest version number for that record.     The difficulty is some records may have multiple version numbers that match the like statement and some may have only one. So this variance with the LIKE search is causing me some confusion.   I've tried this in a few ways using a sub-query but for the life of me I cannot work out how to do it.   Any help would be appreciated,   Drongo  

I'm trying to run a readout of a db which runs fine as an individual script. When I embed it in PHP inside an html div container, it bails when it encounters the first ">" and simply outputs the PHP characters from there through the "?>". The rest of the html runs fine before and after. For example, this line

echo "<select>";

would output

";

and any other php script up to the ?> end, after which it renders html fine. If I run the script as a separate php file, it runs as expected. Any help would be appreciated. Thanks.