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AyKay47

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Everything posted by AyKay47

  1. $c = eval("$b ".$a); would make sense.
  2. I actually forgot that it validates it's contents if required. Here is the regex that it uses for validation: /^[a-zA-Z0-9.!#$%&’*+/=?^_`{|}~-]+@[a-zA-Z0-9-]+(?:\.[a-zA-Z0-9-]+)*$/ not the tightest regex, but efficient enough. and the w3 docs if anyone cares to see: http://dev.w3.org/html5/markup/input.email.html but yah as noted, not all browsers yet support this and server side validation is still required.
  3. my money is on an ambiguity issue with Resource_ID, but add this to the query: $sql="SELECT * FROM resource l inner join resource_skill ln on l.Resource_ID = ln.Resource_ID inner join skill n on ln.Skill_ID = n.Skill_ID WHERE Resource_ID=" . $contactid; //-run the query against the mysql query function $result=mysql_query($sql) or die(mysql_error() . '<br />' . $sql); post the results.
  4. certain browsers will stop parsing when an error is encountered. As the error says, bgcolor is not a valid css property, background-color is. body{background-color:#E0E0E0;} //change accordingly
  5. you can have some sort of validation client side, but should always also have validation server side, just in case JS is disabled, this will prevent security leaks. as jesirose already stated.
  6. yes, you said you are using mysql console vial shell, right?
  7. select col_name from tbl_name where timestamp_field = NOW()
  8. Why would you make such a claim? This is the typical way to go about this.
  9. yah, http://api.jquery.com/attr/
  10. at first glance the script looks to be fine, except you specify q twice in the query string. What happens when the script is ran? Have you tried debugging the script using alert()s?
  11. the first statement, add either a class or an id (depending upon how many images you want the css to affect) and style it in an external css page.
  12. You want JS/Jquery, what have you tried so far?
  13. remove the <br>\n from the previous line. You can add CSS to the image also.
  14. again, this code will need to be tweaked so it looks right: while($row = mysql_fetch_array($result)) { // Loop through results $i++; echo "Displaying record $i<br>\n"; echo "<b>" . $row['id'] . "</b><br>\n"; // Where 'id' is the column/field title in the database echo $row['Location'] . "<br>\n"; // Where 'location' is the column/field title in the database echo $row['Property_type'] . "<br>\n"; // as above echo $row['Number_of_bedrooms'] . "<br>\n"; // .. echo $row['Purchase_type'] . "<br>\n"; // .. echo $row['Price_range'] . "<br>\n"; // .. echo '<img src="'. $row['images'] .'" />'; //image }
  15. Here is some Pseudo code to get you going: $sql = 'SELECT * from table'; $query = mysql_query($sql) or die(mysql_error() . '<br />' . $sql); while($row = mysql_fetch_assoc($query)) { ?> <img src="<?php echo $row['images']; ?>" /> <?php } this will simply display each image in the table, you will need to tweak it to suit your needs.
  16. looks right. Now you need to grab that data from the db using a select statement. Have you attempted a query yet?
  17. create a varchar type field named whatever and insert the file paths. So if you have an image at /some/file/path/image.jpg, that is what will go into that field, along with any other corresponding image data that you want.
  18. Typically, the images are stored in a directory and the file paths are stored in the database.
  19. the query that you are providing mysql_fetch_array() is returning a boolean false. The problem lies there.
  20. you misread, I said it shouldn't be there.
  21. Nice layout, nice text colors, overall a nice site. 1. The green top nav buttons seem a little random to me, doesn't really go with the color scheme imo. 2. maybe change the top user/pass inputs from the default outline. I will update my list if I spot anything else. This is a good looking site.
  22. The data needs to be stored somewhere. How will there be 120 users if the data is not stored?
  23. It should be. How is the data saved? Please don't say to a .txt file.
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