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matthew9090

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Posts posted by matthew9090

    my forum

    in Website Critique

    Posted

    yes, my hosting is free so it is down sometimes.

     

    This site is a test to see if it is a success, otherwise if there is no progress i will think of another idea. I really want a business when im older and i want to work from home. So im already starting :).

    code tidy

    in PHP Coding Help

    Posted

    please could you inspect this code and tell me any errors. It is a dictionary script so i shows words from a database with the definition underneith.

     

    here are ones that i can't fix:

    • There are 2 divs ontop of each other which makes some borders thicker than the other
    • Some searches start out as bold word and non-bold descripton(that's meant to happen) but then half way down all the text is bold and the borders are thicker
    • the extend() javascript function does not work

     

    other than that everything else works.

     

    so please can you help clean up my code and tell me what i need to do to fix some of the things.

     

    here's the code:

    <style type="text/css">
      div 
      {
      font-family:Arial,Helvetica,sans-serif;
      font-size: 30px;
      border-style:solid;
border-width:1px;
width: 700px;
background-color: #2CFFB5;
padding: 0px;
      }
      div#div:hover {
background-color:#f1f1f1;
cursor:pointer
      }
      .highlight {
      color: red;
      }
</style>
<script type="text/javascript">
function hover() 
{
document.getElementById("div").style.border="thick solid #0000FF";
}

function extend(str)
{
document.getElementById("extend").innerHTML=str;
document.getElementById("extend").style.width="800px";
}
</script>
<?php
error_reporting(E_ALL);
mysql_connect("localhost", "root", "");
mysql_select_db("dictionary");
$words = mysql_query("SELECT * FROM words");
$q = strtolower($_GET['q']);
$query = mysql_query("SELECT * FROM words WHERE word LIKE '%$q%' OR definition LIKE '%$q%'");
while ($row = mysql_fetch_assoc($query))
{
   $tags1 = "<div id=\"div\" onMouseOver='hover()'><b>";
   $tags2 = "</b><br />";
   $tags3 = "<br /><br /></div>";
   $string1 = $row['word'];
   $string2 = $row['definition'];
   if (strlen($string2)>43) 
   { 
      $tags1 .= "<div id='extend' onMouseOver=\"extend(".$string2.")\">" . $tags1;
      $tags3 = $tags3 . "</div>";
      $string2 = substr($string2, 0, 40); 
      $tags3 = "...<br /><br /></div>";
   }
   $string3 = str_ireplace($q, "<span class='highlight'>$q</span>", $string1);
   $string4 = str_ireplace($q, "<span class='highlight'>$q</span>", $string2);
   $string = $tags1 . $string3 . $tags2 . $string4 . $tags3;
   $string = strtolower($string);
   if (strlen($string)>25) 
   { 
      substr($string, 0, 25); 
   }
   echo $string;
}

if (mysql_num_rows($query)==0)
{
   echo "No words found";
}

if ($q=="")
{
   echo "Type a word";
}
?>

     

    and the index page:

    <html>
   <head>
      <script type="text/javascript" src="scripts.js"></script>
      <style type="text/css">
      p {
      font-family:Arial,Helvetica,sans-serif;
      color: #FF4B2C;
      }
      
      a:link {
      color: #FF4B2C;
      }
      
      a:visited {
      color: #FF4B2C;
      }
      
      html {
         background-color: #2CE0FF;
      }
      
      h1 {
      color: #2C76FF;
      }
      
      #input {
         width: 700px;
         height: 55px;
         font-family: verdana;
         font-size: 35px;
         background: #A4EF9C url(a.png) no-repeat left -15px ;
      }
      #input:hover {
         background-color: #ffffff;
      }
      #input:focus {
         background: #ffffff url(a.png) no-repeat left -11125px ;
      }
      
      h1 {
         font-size: 50px;
      }
      
      div {
      
      }
      </style>
   </head>
   <body>
      <h1>The dictionary</h1>
      <form>
         <input type="text" id="input" onkeyup="showHint(this.value)" /> 
      </form>
      <p>
      <div id="txtHint"></div>
      </p>
      <p>
         <a href="word.php">Make up your word now!</a>
      </p>
   </body>
</html>

 

     

    and the ajax script:

    function showHint(str)
{
if (str.length==0)
  { 
  document.getElementById("txtHint").innerHTML="";
  return;
  }
if (window.XMLHttpRequest)
  {// code for IE7+, Firefox, Chrome, Opera, Safari
  xmlhttp=new XMLHttpRequest();
  }
else
  {// code for IE6, IE5
  xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
  }
xmlhttp.onreadystatechange=function()
  {
  if (xmlhttp.readyState==4 && xmlhttp.status==200)
    {
    document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
    }
  }
xmlhttp.open("GET","gethint.php?q="+str,true);
xmlhttp.send();
}

    php live search

    in PHP Coding Help

    Posted

    i prefere to do it my self and i have got this so far:

     

    <?php
error_reporting(E_ALL);
mysql_connect("localhost", "root", "");
mysql_select_db("dictionary");
$words = mysql_query("SELECT * FROM words");
$a = array();
$count = 0;
while ($row = mysql_fetch_assoc($words))
{
   $a[$count][] = $row['word'];  
   $a[$count][] = $row['definition']; 
   $count++;
}
$q = $_GET['q'];
if (strlen($q) > 0)
  {
  $hint="";
   foreach ($a as $entry){
   if (substr($a[$count][0],0,strlen($q))) //line 19<<<<<<<<<
      {
      if ($hint=="")
        {
          $hint= "<strong>" . $a[$count][0] . "</strong><br /><em>" . $a[$count][1] . "</em>"; 
        }
      else
      {     
        $hint = $hint . "</b><br /><br /> <strong>" . $a[$count][0] . "</strong><br /><em>" . $a[$count][1] . "</em>"; 
        }
      }
    }
}

// Set output to "no suggestion" if no hint were found
// or to the correct values
if ($hint == "")
  {
  $response="no suggestion";
  }
else
  {
  $response=$hint;
  }

//output the response
echo $response;
?>

     

    and i get the errors:

    Notice: Undefined offset: 5 in E:\wamp\www\dictionary\gethint.php on line 19

     

    Notice: Undefined offset: 5 in E:\wamp\www\dictionary\gethint.php on line 19

     

    Notice: Undefined offset: 5 in E:\wamp\www\dictionary\gethint.php on line 19

     

    Notice: Undefined offset: 5 in E:\wamp\www\dictionary\gethint.php on line 19

     

    Notice: Undefined offset: 5 in E:\wamp\www\dictionary\gethint.php on line 19

     

    Allowing a comma in the price field

    in PHP Coding Help

    Posted

    Hi Matthew Not sure how to incorporate that with my code.. Please advise

    function check_Price($price,$member_id,$description,$ip){
if (isset($price) && (!is_numeric($price)))
{
      setlocale(LC_MONETARY, 'en_US');
      echo money_format('%i', $price) . "\n";
}

}

     

    should work

     

    php live search

    in PHP Coding Help

    Posted

    im am creating a live search using the w3schools tutorial. apart from im using mysql.

     

    here is my current code(which works fine):

     

    <?php
mysql_connect("localhost", "root", "");
mysql_select_db("dictionary");
$words = mysql_query("SELECT * FROM words");
$a = array();
while ($row = mysql_fetch_assoc($words))
{
   $a[] = $row['word'];  
}
$q = $_GET['q'];
if (strlen($q) > 0)
  {
  $hint="";
  for($i=0; $i<count($a); $i++)
    {
    if (strtolower($q)==strtolower(substr($a[$i],0,strlen($q))))
      {
      if ($hint=="")
        {
        $hint=$a[$i];
        }
      else
      {     
        $hint = $hint . "</b><br /><br /> ".$a[$i];
        }
      }
    }
  }

// Set output to "no suggestion" if no hint were found
// or to the correct values
if ($hint == "")
  {
  $response="no suggestion";
  }
else
  {
  $response=$hint;
  }

//output the response
echo $response;
?>

     

    as you can see it is edited alot. But my problem is that i need to display underneath the word, the definition aswell. I've been trying to figure this out but i can't seem to do it. I have been trying to use multidimensional arrays.

    Echoing an HTML Tag...

    in PHP Coding Help

    Posted

    put a backslash (\) before every quote

     

    like this:

     

    echo '<td width=\"300\" bgcolor=\"#ffffff\" align=\"center\" onmouseout=\"style.backgroundColor=\'#ffffff\'\" onmouseover=\"style.backgroundColor=\'#e8e8e8\';\" style=\"word-wrap: break-word; border: 1px solid rgb(238, 238, 238); font-size: 12px; background-color: rgb(255, 255, 255);\">\';

    queston

    in PHP Coding Help

    Posted

    this is hard to explain but how can you make it so as php executed the script it shows whats happening as it goes. An example of this is the google sitemap generator. it shows the url every time it finds a new one. Do you need to use ajax.

    oop help

    in PHP Coding Help

    Posted

    i was just testing out my oop and wrote this myself

     

    <html>
<body>
<?php
error_reporting("E_ALL");
   class show {
      public function connect() {
         mysql_connect("localhost", "root", "");
      }
      public function sel_db() {
         mysql_select_db("oop");
      }
      public function get() {
         $query = mysql_query("SELECT * FROM ppl");
         while ($row = mysql_fetch_array($query)) 
         {
            echo $row['id'] . " | " . $row['name'] . "<br />";
         }
         
         $num_rows = mysql_num_rows($query);
         echo $num_rows;
      }
   }
   
$a = new show;

$a->connect;
$a->sel_db;
$a->get;

?>
</body>
</html>

     

    but the query doesn't work. I don't get any errors. It will not show anything on the page. It is supposed to show all results from a table.

    link scrolling program

    in Miscellaneous

    Posted

    im currently using a program called Xenu to browse all of my websites links and report any broken ones.

     

    im just wondering if there is a program that doesn't just scroll the links from your site, but then keeps scrolling the other sites found.

    run form on page load

    in HTML Help

    Posted

    this might work

     

    <form method="post" id="myForm" action="someotherpage.php">

    <input type="text" name="name" value="Value" />

        <button type="submit" onload="this.form.submit()">Submit</button>

    </form>

    PHP Redirection, Youtube fault.

    in PHP Coding Help

    Posted

    if you want to check if the url is valid then do something like this:

     

    <?php
$url = $_SERVER['HTTP_HOST']; //get domain name of the url submitted

if (stristr($url, "http://"))
{
    //now do the same for www or co.uk, .com etc......
}
else
{
die("url is not valid");
}
?>

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