h0m3r0w Posted April 15, 2013 Share Posted April 15, 2013 Hello all, I'm having a bit of a problem with my code here. I'm trying to allow the user to enter a number (an office number) from a database, and based on that value, after submitting it, an html table will fill with the appropriate results (lastName, firstName, jobTitle). For whatever reason, though, I keep getting this error: "Warning: mysqli_fetch_array() expects parameter 1 to be mysqli_result, boolean given in... <website name> <line number>" It's very weird, because I'm doing this straight from notes / w3 and I'm still getting errors. I'm still new to this, so I'm sorry if I'm sounding completely inexperienced... well I am haha. Anyway, here's the code: <!DOCTYPE html > <html lang ="en"> <head> <title> Homepage </title> </head> <body bgcolor="gray"> <h1 align=middle> Welcome to the Site </h1> <?php DEFINE ('DB_USER', '*******); DEFINE ('DB_PASSWORD', '******'); DEFINE ('DB_HOST','*****'); DEFINE ('DB_NAME','******'); $dbc = @mysqli_connect(DB_HOST,DB_USER,DB_PASSWORD,DB_NAME) or die ('Could not connect to MySQL: '.mysqli_connect_error($dbc) ); mysqli_set_charset($dbc, 'utf8'); //$r = @mysqli_query($dbc, $q); ?> <form action ="Homepage.php" method="get"> Enter Office Code: <input type="text" name="oCode" id="OfficeCode" maxlength="15"> <input type="submit" name="formSubmit" value="Submit"> </form> <? $code = $_GET['oCode']; $result = mysqli_query($dbc, "select lastName, firstName, jobTitle from Employees where OfficeCode='$code'"); echo "<table border='1'>"; echo "<tr>"; echo "<th>LastName</th>"; echo "<th>FirstName</th>"; echo "<th>Job Title</th>"; echo "</tr>"; while ($row = mysqli_fetch_array($result)) //this is where I'm getting the error { echo "<tr>"; echo "<td>" . $row['lastName'] . "</td>"; echo "<td>" . $row['firstName']. "</td>"; echo "<td>" . $row['jobTitle'] . "</td>"; echo "</tr>"; } echo "</table>"; mysqli_close($dbc); ?> Any help would be great. Thank you! Link to comment https://forums.phpfreaks.com/topic/276992-filling-php-table/ Share on other sites More sharing options...
lemmin Posted April 15, 2013 Share Posted April 15, 2013 I think you have an error in your query. Try using mysqli_error() to find out what it is: $q = "select lastName, firstName, jobTitle from Employees where OfficeCode='$code'" if (!($result = mysqli_query($dbc, $q))) error_log(mysqli_error($dbc)."<br/>\nQuery: ".$q); Link to comment https://forums.phpfreaks.com/topic/276992-filling-php-table/#findComment-1424959 Share on other sites More sharing options...
computermax2328 Posted April 15, 2013 Share Posted April 15, 2013 I think I see the problem or part of the problem. It is telling you that the first parameter of your mysqli_query argument is incorrect, so the $dbc is not working. Look at this... DEFINE ('DB_USER', '*******); DEFINE ('DB_PASSWORD', '******'); DEFINE ('DB_HOST','*****'); DEFINE ('DB_NAME','******'); Your DEFINE DB_USER is missing a ' at the end of the ********. See if that works! Link to comment https://forums.phpfreaks.com/topic/276992-filling-php-table/#findComment-1424964 Share on other sites More sharing options...
h0m3r0w Posted April 15, 2013 Author Share Posted April 15, 2013 On 4/15/2013 at 10:29 PM, lemmin said: I think you have an error in your query. Try using mysqli_error() to find out what it is: $q = "select lastName, firstName, jobTitle from Employees where OfficeCode='$code'" if (!($result = mysqli_query($dbc, $q))) error_log(mysqli_error($dbc)."<br/>\nQuery: ".$q); Here, I noticed that I'm a complete idiot. I forgot to select a specific database... When I connect to mySQL, I have a list of databases. While the one I hidden was indeed a database, it wasn't the correct one... So, now that I have that done, I'm still getting the exact same error for some reason... On 4/15/2013 at 11:01 PM, computermax2328 said: I think I see the problem or part of the problem. It is telling you that the first parameter of your mysqli_query argument is incorrect, so the $dbc is not working. Look at this... DEFINE ('DB_USER', '*******); DEFINE ('DB_PASSWORD', '******'); DEFINE ('DB_HOST','*****'); DEFINE ('DB_NAME','******'); Your DEFINE DB_USER is missing a ' at the end of the ********. See if that works! Haha, sorry but that was a typo on my part. There actually was another single quote, I just made an error while trying to conceal some of the information. I'm no good at PHP, so I'm unsure as to how much code I should be revealing, since I'm connecting to my network and all that. But all in all, I made a dumb error in connecting to the wrong database. But even with the correct database selected, I get an error. Well, I BELIEVE that I'm connected to the database. Do I need to use a "use ********" statement in the php? Where the **** is the name of one of the databases ("classicmodels") that I need to connect to? Link to comment https://forums.phpfreaks.com/topic/276992-filling-php-table/#findComment-1424970 Share on other sites More sharing options...
jcbones Posted April 16, 2013 Share Posted April 16, 2013 No, just make sure your database name is filled out in the DB_NAME constant definition. It is being passed to the mysqli class in the constructor. What is the error you are getting now? Link to comment https://forums.phpfreaks.com/topic/276992-filling-php-table/#findComment-1424979 Share on other sites More sharing options...
h0m3r0w Posted April 16, 2013 Author Share Posted April 16, 2013 Well, on that note, I have gotten everything completed. Thank you guys so much for the help. So, at the end of the day, it was me inserting the incorrect database... Thanks again guys! Link to comment https://forums.phpfreaks.com/topic/276992-filling-php-table/#findComment-1424980 Share on other sites More sharing options...
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