mysqli query

[urielk]

Hi,

 

I am having problems returning values from a select statement. When I query directly in the databse, I get back the information I am looking for. I use an includes file for the database connection and my page shows that the connection was successful. Here is my code:

 

<?php
  
    $search = $_GET['searchFor'];
    $words = explode(" ", $search);
    $phrase = implode("%' AND articlename LIKE '%", $words);

   	$sql ="SELECT * FROM articles WHERE articlename LIKE '%phrase%'";
    $result =$conn->query($sql) or  die('Sorry, could not get any articles at this time');
    $row =($result->fetch_all()) or die('No records found');
    $numRows =$result->num_rows;

    If($numRows==0)
    {

    echo "<h2>Sorry, no articles were found with '$search' in them.</h2>";
    } 
else
    {
        While($row=$result->fetch_assoc())
        {
            $articleid = $row['articleid'];
            $title = $row['articlename'];
            $shortdesc = $row['shortdesc'];
		echo "<h2>Search Results</h2><br><br>\n";
            echo "<a href=\"index.php?content=showarticle&id=$articleid\">$title</a><br>\n";
            echo "$shortdesc<br><br>\n";
        }
    }
?>

 

The search term is coming from a search form in the navigation. I have used "echo" statements to check and make sure that the sesrch word is coming through to tghe page containing the above code. I have tried mysqli_error() statements in several places and don't see where the problem is. When I try the search the message that comes back is "No records found" Does not makee sense because I know it is there, can find it, and even have the same syntax as the SELECT statement I use when I ask for the php code. Going crazy trying to sort this out. Any suggestions, help etc are greatly appreciated. Thank youi.

[Pikachu2000]

See anything wrong here?

$phrase = implode("%' AND articlename LIKE '%", $words);

      $sql ="SELECT * FROM articles WHERE articlename LIKE '%phrase%'"; // <---------- HERE . . .

[urielk]

New to php so I apologize but I don't see an error in the statement. I looked at hph code hint by doing a query using phpMyAdmin. They seemed to have a parentheses between Where and the before the double quotes. I tried that and the query says there are no records when I can find them. For some reason the query is not picking up the $phrase and I am n0ot sure how to figure out why.  Be a big help if anyone can point me to a way to get this information. Thanks.

[Pikachu2000]

You implode the $words array and assign it to the variable $phrase, then you never use $phrase in the query string. Right now the query will only match records that actually have the word "phrase" in the articlename field.

[urielk]

Hi,

 

Thank you for pointing out the error. Did not see it neither did someone else who looked at the code several times. I made the change but I am still not getting any results from the query. I have a file index.php where I have an included file for the connection. I also have an included file for the sidebar which is where I have the form for the search. I left an echo statement in my connection file to tell me if the connection is successful and that statement is still there and I get a message that the connection is successful. My query is not working and I am not sure what statements to use to find out the source of the error so I can fix it. Thanks for all the help.

[mikosiko]

do a simple test...

 

  $sql ="SELECT * FROM articles WHERE articlename LIKE '%phrase%'"; // assuming that you already changed this line as Pica suggested

 

  echo "** The SQL is : " . $sql;  // add this line and post what your get

 

  $result =$conn->query($sql) or  die('Sorry, could not get any articles at this time');

 

also.. is good idea to include this 2 lines at the beginning of your code for debugging:

error_reporting(E_ALL); 
ini_set("display_errors", 1);

[urielk]

Hello Everyone,

 

Thanks for the patience. I made the correction suggested by Pica and also added both the error reporting and ini_set statements. What I get back is:

 

** The SQL is : SELECT * FROM articles WHERE articlename LIKE '%Metabolism%'

 

That is all I get and I copied it exactly as it appears. So if the search term is being passed to the query and I can find the result from phpMyAdmin but not from the program, where are things going wrong?  Thanks for the help.

[mikosiko]

if you echo $numRows  what do you get?

[urielk]

I get a "1" in addition to the statement that echoed back earlier. Here is an exact copy:

 

** The SQL is : SELECT * FROM articles WHERE articlename LIKE '%Metabolism%'1

 

OK so it returns one row which is non-zero. Since the if statement is false it should go to else and execute the while loop right? Do I need a count at the end of the query so it has some limit to step through? Maybe that is the problem. Thanks for any additional comments or suggestions.

[mikosiko]

check your logic again....

 

in your code:

 $row =($result->fetch_all()) or die('No records found');
    $numRows =$result->num_rows;

    If($numRows==0)
    {

    echo "<h2>Sorry, no articles were found with '$search' in them.</h2>";
    } 
else
    {
        While($row=$result->fetch_assoc())

 

you first

$row =($result->fetch_all()) or die('No records found');

already got the first record... which happens to be the only one, and moved the internal pointer... therefore your while loop is no processing anything... is nothing to process right?

[urielk]

I am not sure I agree. A similar code in another program seems to work fine and even though there is only one result, it returns the details - articlename, date, and short description - where the article name becomes a hyperlink allowing one to get the full article text. That is what I am trying to do here and it is not working. That is the reason for my request. Later as I add more articles I am expecting the program to return a hyperlinked list so people get the details. Thanks for your comment anyway.