PHP - I Want To Search More In Like Clause..

Hi,

I have a SQL statement:
$sql='SELECT PROPID, ADDRESS_1, ADDRESS_2, TOWN, POSTCODE1, POSTCODE2, BEDROOMS, BATHROOMS, RECEPTIONS, PRICE, TRANS_TYPE_ID FROM properties WHERE TRANS_TYPE_ID=2 AND BEDROOMS =3 AND PROP_SUB_ID IN (1,2,3,4,5,6,21,22,23,24,27,30,95,128,131) ORDER BY DATE_ADDED, PROPID DESC';

I want to be able to get the fields and their corresponding values as such:

$result['TRANS_TYPE_ID']=2;
$result['BEDROOMS']=3;

And then capture the values in PROB_SUB_IN IN(*)

How can I do this?

Any help will be greatly appreciated. Thanks

Hi,

I have an empty array and i populated the array through code and i printed to check if everything ok. So far so good. I now have an array say $ids = ('123','456'). Now, i have a table in db that has ids as well, I need to make a query to check for id's that are not inside the array, so if the db table have '123','456','789' i want the result of the query to be only the last one '789'.

$arr = array();
array_push($arr,$previd2);
print_r ($arr);

I get the result as follows:

Array ( [0] => 533349 [1] => 533355 )

so what query i can use to get the other id's from db table that is not part of that array?

here is what i tried:

$qry = "select staffname, joindate from staff WHERE OracleID NOT IN ($arr)";
$answ = mysqli_query($con, $qry);

I get error:

Notice
: Array to string conversion in
C:\xampp\htdocs\AttendanceSystem\login\reportsforallid.php

 

Edited March 30, 2020 by ramiwahdan
more info

I have this code:


Code: [Select]
if (isset($_POST['update'])){
$ids = implode(",", array_map('intval', $_POST['m']));
$ranks = implode(",", array_map('intval', $_POST['ranks']));
if ($ids < 1)
message("Incorrect Data");
if ($ranks < 0)
message("Incorrect Data");
echo $ids;
echo "<br>";
echo $ranks;
exit;
$db->query('UPDATE friends set RANKS WHERE friend_id IN ('.$db->escape($ids).') AND user_id = '.$pun_user['id'].'') or error('Unable to remove users from online list', __FILE__, __LINE__, $db->error());
redirect("s.php?section=Friends","Thanks, Ranks Updated");
}
I grab the id's from the rank

This will spit out 3,4 for $ids and for $ranks, 1,2 as you can see, set for 1 and 2 respectivaly.

My problem is how can I use MYSQL to update the data correspond to each id using a IN clause?

Code: [Select]
$db->query('UPDATE friends set RANKS = "LOST $RANKS? HERE" WHERE friend_id IN ('.$db->escape($ids).') AND user_id = '.$pun_user['id'].'') or error('Unable to remove users from online list', __FILE__, __LINE__, $db->error());

Hi,   Im having a problem and I can't seem to figure it out or find anything on the net.   If I use the following code the script successfully updates every row in the table:

mysqli_query($con,"UPDATE Ads SET Ads_LocalArea='Stroud'");

However if I try updating the table using the WHERE clause in any of the combinations below nothing happens.

mysqli_query($con,"UPDATE Ads SET Ads_LocalArea='Stroud' WHERE Ads_ID=$DBROWID");

----------------------------------------------------------------------   My Script:

mysqli_query($con,"INSERT INTO Ads (Ads_ID, Ads_AID, Ads_Title)
VALUES ('', '$Polished_AdRef', '$Polished_AdTitle')");

$DBROWID = mysqli_insert_id($con);

mysqli_query($con,"UPDATE Ads SET Ads_AID='Stroud' WHERE Ads_ID=$DBROWID");

// TRIED THESE TOO
// mysqli_query($con,"UPDATE Ads SET Ads_AID='Stroud' WHERE Ads_ID='$DBROWID'");
// mysqli_query($con,"UPDATE Ads SET Ads_AID='Stroud' WHERE Ads_ID='5'");

Does any one know where I am going wrong?

Hey guys,

I don't know what to do. I've got a script used for gamble:


$gamble_rand = rand($number_of_tickets,$gamble_chance);
if($gamble_chance == $gamble_rand){


If $number_of_tickets is set by 99 and $gamble_chance is set by 100 the chance is 1:100 that the If-Clause is not true. But it seems to me that PHP is making its on "random". Independent of the values PHP is interpreting that 20 times of 100 times the correct logic is false...

If $number_of_tickets is set by 1 there's no big different. Its just a little bit less, but not the chance that it has to be.

Thanks for helping and Merry Christmas :-)

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=330779.0

hello.

i have a column called "pageName" and on the each page i have put $pName = "the name of that page".

i want to pull from the database all the information where the $pName on the page is the same as the pageName in the database

its getting the $pName = "adminHome" but it thinks its the name of the column not the row.

this is the error im getting:

Quote

DATABASE.PHP - confirm_query = MySQL Datatbase Query Failed: Unknown column 'adminHome' in 'where clause'

Last SQL query: SELECT * FROM pages WHERE pageName=adminHome



this is the function code
Code: [Select]
public static function find_by_pageName($pName=""){
global $database;
$sql = "SELECT * FROM ".self::$table_name." WHERE pageName=".$database->escape_value($pName)."";
$result_array = self::find_by_sql($sql);
return !empty($result_array) ? array_shift($result_array) : false;
}


this is the page code
Code: [Select]
<?PHP
require_once("../includes/initialize.php");

$pName = "adminHome";

$page = Pages::find_by_pageName($pName);
?>



thanks
rick

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=350684.0

Hi,

I've been playing around with inserting the name of a file ($additional_upload_nameX)  into the database only if it has been uploaded (it's not mandatory).

When I submit the form, I'm told I have a problem with my WHERE clause. Any idea what I'm doing wrong here?

Code: [Select]
$query2 = "INSERT INTO uploads (date, upload_name, topic, year, status, keywords, description )" . "VALUES (NOW(), '$additional_upload_nameX', '$topic', '$year',  '$status', '$keywords' , '$description') WHERE $additional_upload_nameX != ''";
//if($query){echo 'data has been placed'}

mysql_query($query2) or die(mysql_error());

I have a table which contains a TINY INT column. If there are images associated with the post, the TINY is 1. If not, 0.   I want to order results by images first, and earliest DATETIME first.   Basically, like using a boolean but not. Simply ordering by the TINY INT column ASC or DESC isn't working. Not sure how to solve this. Help appreciated. Thanks!

I had originally created a topic on my pagination not working, but I found out what the problem was. The topic I had created was irrelevant to what the problem was... but anyway, here's my situation:

I've got a query that searches in a MySQL table for a certain value. Then it list's all the findings in a table (pretty simple). Anyway, I obviously want to have a pagination system to make it easier to manage the amount of data per page.

The pagination works fine, it's used for another query that just pulls data (so NO WHERE clause is used!). That is what is causing the problem. I spent hours trying to figure out why this is happening, then I figured out that the only difference between the pagination that works and the one that doesn't is the query have the where clause.

What happens when the where clause is used is that the pagination works fine on the first page, but if you click the link to go to a different page, the page is empty and turns up no results so the mysql data that is pulled is lost.

So here is my code. I don't know exactly why the where clause causes the pagination to not work.

Code: [Select]
$conn = mysql_connect("$mysql_host","$mysql_user","$mysql_password") or trigger_error("SQL", E_USER_ERROR);
$db = mysql_select_db("$mysql_database",$conn) or trigger_error("SQL", E_USER_ERROR);

// find out how many rows are in the table
$sql = "SELECT COUNT(*) FROM ACTIVE WHERE MODEL='$model'";
$result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR);
$r = mysql_fetch_row($result);
$numrows = $r[0];

// number of rows to show per page
$rowsperpage = 5;
// find out total pages
$totalpages = ceil($numrows / $rowsperpage);

// get the current page or set a default
if (isset($_GET['currentpage']) && is_numeric($_GET['currentpage'])) {
   // cast var as int
   $currentpage = (int) $_GET['currentpage'];
} else {
   // default page num
   $currentpage = 1;
} // end if

// if current page is greater than total pages...
if ($currentpage > $totalpages) {
   // set current page to last page
   $currentpage = $totalpages;
} // end if
// if current page is less than first page...
if ($currentpage < 1) {
   // set current page to first page
   $currentpage = 1;
} // end if

// the offset of the list, based on current page
$offset = ($currentpage - 1) * $rowsperpage;


$sql = "SELECT * FROM ACTIVE WHERE MODEL='$model' ORDER BY INDEX_ID DESC LIMIT $offset, $rowsperpage";
$result = mysql_query($sql, $conn) or die('Error: ' . mysql_error());

echo "
<table width=100% align=center border=1 class=\"mod\"><tr>
 <td align=center bgcolor=#00FFFF><b>Rating</b></td>
 <td align=center bgcolor=#00FFFF><b>Title</b></td>
 <td align=center bgcolor=#00FFFF><b>Make/Model</b></td>
 <td align=center bgcolor=#00FFFF><b>Type</b></td>
 <td align=center bgcolor=#00FFFF><b>Difficulty</b></td>
 <td align=center bgcolor=#00FFFF><b>Comments</b></td>
 <td align=center bgcolor=#00FFFF><b>Views</b></td>
</tr>";
   
     while ($row = mysql_fetch_assoc($result)) {
     { // Begin while
      $title = $row["TITLE"];
      $type = $row["TYPE"];
      $difficulty = $row["DIFFICULTY"];
      $comments = $row["COMMENT_TOTAL"];
      $id = $row['INDEX_ID'];
      $rating = $row["RATING_AVERAGE"];
      $make = $row["MAKE"];
      $model = $row["MODEL"];
      $views = $row["HITS"];

     echo "
     <tr>
      <td>$rating/10</td>
      <td><a href=\"mod.php?id=$id\">$title</a></td>
      <td>$make - $model</td>
      <td>$type</td>
      <td>$difficulty</td>
      <td>$comments</td>
      <td>$views</td>
     </tr>";
    }
   }
   echo "</table><br />";
    /******  build the pagination links ******/
echo "<center>";
// range of num links to show
$range = 3;
echo "<i>Page: </i>";
// if not on page 1, don't show back links
if ($currentpage > 1) {
   // get previous page num
   $prevpage = $currentpage - 1;
   // show < link to go back to 1 page
   echo " <a stylehref='{$_SERVER['PHP_SELF']}?currentpage=$prevpage'>Previous</a> - ";
} // end if

// loop to show links to range of pages around current page
for ($x = ($currentpage - $range); $x < (($currentpage + $range) + 1); $x++) {
   // if it's a valid page number...
   if (($x > 0) && ($x <= $totalpages)) {
      // if we're on current page...
      if ($x == $currentpage) {
         // 'highlight' it but don't make a link
         echo " <b>$x</b> ";
      // if not current page...
      } else {
         // make it a link
         echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=$x'>$x</a> ";
      } // end else
   } // end if
} // end for
                 
// if not on last page, show forward and last page links       
if ($currentpage != $totalpages) {
   // get next page
   $nextpage = $currentpage + 1;
    // echo forward link for next page
   echo " - <a href='{$_SERVER['PHP_SELF']}?currentpage=$nextpage'>Next</a> ";
} // end if
/****** end build pagination links ******/
mysql_close($conn);
}
echo "$pagination</center>";
?>

I really appreciate the help! Thanks.

Hi, I'm making an attempt to build a small cms based site. I have read in a lot of places it is best to use the id of a table row to fetch contents, but being as I am using url rewrite in frontend, it won't be showing query strings, so I thought I would ask if anyone saw any potential problems if I did the following.

Page table looks like this:
Pages
- id
- name
- content

In backend, the user has a form to name the page and use ckeditor to add rich content. I will use php to ensure that the name is alphanumeric and use strtolower & str_replace to ensure that My First HTML Page is sent to mysql as my-first-html-page and also check that a row with this name does not already exist.
So on front end this page url will be mywebsite.com/my-first-html-page

So to output the content I strip out the domain & / so I'm left with the name as entered in the db.

$url = $_SERVER['REQUEST_URI'];
$url = str_replace ('/','',$url);

Then in my function to output content I will use WHERE name = $url

This topic has been moved to MySQL Help.

http://www.phpfreaks.com/forums/index.php?topic=332417.0

I'm a newbie, can you please help?

I'm getting a mysql error "Unknown column 'E0000001' in 'where clause'"

The id is in the URL: ...user-profile.php?id=E0000001

My Query

$query = mysqli_query($con, "SELECT * FROM UserList WHERE UserID=".$id) or die (mysqli_error($con));

Thank you

Edited September 18, 2019 by ACBMSE