CODEWITHSUNDEEP
phpmysqliswitch-statement
user995426
user995426

Reputation: 91

Issue with PHP and mysql using SELECT and switch

Hi I am trying to Select table based upon directory structure but query is not working properly. Pets is main folder inside which there are folders Dogs , Cats , fishes, Horses http://animalswecare.com/Pets/Dogs/

http://animalswecare.com/Pets/Cats/

http://animalswecare.com/Pets/Horses/

I am selecting tables based upon dirctory structure like if page is under Pets>Dogs then table dogs_DB , if same code or page is put in Pets>Cats then table cats_DB I have used echo for checking and output is at the end of code.

 <?php
include('../../connectuser.php');
echo $ab=$_SERVER['PHP_SELF'];
$show=explode ("/",$ab);
echo $show[0]."<br />";
echo $show[1]."<br />";
echo $show[2]."<br />";
echo $pagename= $show[1].$show[2];
echo "<br />";
switch ( $pagename ) {

 case 'PetsDogs':  
 $tbl_name = 'dogs_DB';
 break;

 case 'PetsCats':  
 $tbl_name = 'cats_DB';
 break;

 case 'PetsFishes':  
 $tbl_name = 'fishes_DB';
 break;

 case 'PetsHorses':  
 $tbl_name = 'horses_DB';
 break;
 }
 echo $tbl_name;
 echo "<br />";
 echo $sel = "SELECT * FROM $tbl_name" or die('error in query');
 $fetch = mysqli_query( $dbc, $sel );

 while ( $row = mysqli_fetch_array( $fetch ) ) {

 echo $title;
 $title = $row['title'];
 }
 echo "<br />";

  echo $title."hiii";
 ?>

http://animalswecare.com/Pets/Dogs/get_links.php

Output when page is put under Pets>Dogs folder /Pets/Dogs/get_links.php

Pets

Dogs

PetsDogs

dogs_DB

1

hiii

Upvotes: 0

Views: 176

Answers (4)

Phebus40
Phebus40

Reputation: 173

$title seems to be null at the end of the loop while ( $row = mysqli_fetch_array( $fetch ) )

Upvotes: 0

Alex L
Alex L

Reputation: 390

shouldn't these two lines be the other way around?

echo $title;
$title = $row['title'];

so:

while ( $row = mysqli_fetch_array( $fetch ) ) {

    $title = $row['title'];
    echo $title;
}
echo "<br />";

echo $title."hiii";

Upvotes: 0

Xatep
Xatep

Reputation: 353

Initiate a variable like

$pagename = $show[1].$show[2];

NOT

echo $pagename= $show[1].$show[2];

The same for the $title variable inside the while-loop

echo $title;
$title = $row['title'];

Do instead

$title = $row['title'];
echo $title;

Since it looks like you want to use the $title variable outside the loop, you will also need to declare it outside.

$title = "";
while ( $row = mysqli_fetch_array( $fetch ) ) {
 $title = $row['title'];
 echo $title;     
}
echo "<br />";
echo $title."hiii";

Upvotes: 0

max
max

Reputation: 2817

It seems that the query row with $row['title'] set to empty string. If it's null, the last echo won't output anything. Try var_dump() to get more information about variable.

Upvotes: 1

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