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  1. #1
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    could not print out value

    Hi;

    My problem is i could not print out value after i inserted echo "<div class=\"hints\">; and echo "</div>"; please see the highline .

    I am implementing an auto suggestiion for search engin.

    response.php

    PHP Code:
    <?php
     $q
    =$_GET["q"];
     include(
    "../condatabase.php"); 

     
    $sql="SELECT content FROM articles WHERE content like '%".$q."%'";
     
     
     
    $result mysql_query($sql);

      
    $a[]=""$m[]="";



       while(
    $row mysql_fetch_array($result)){
        
    $m.= $row['content'];
       }
      
    //$final = preg_match_all("/talking a[a-zA-Z]*/", $string);
      
    $term="/" $q "[a-zA-Z\s]*" "/";

      
    preg_match_all($term$m,$f);

     
     
     
    $count=count($f[0]);
     
    $f[0]=array_unique($f[0]);


     
    $i=0;
     while(
    $i<$count){
       
    $a[$i]=($f[0][$i]);
     echo 
    "<div class=\"hints\">;
      echo  $a[$i];  // i could not print out this value, but it can do without two div
     echo "
    </div>";
      $i++;
     }

    ?>
    -------------------------------------------------------------------
    index.php

    PHP Code:
    <html>
    <
    head>
    <
    script type="text/javascript">
    function 
    showHint(str)
    {

    var 
    xmlhttp;
    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==&& xmlhttp.status==200)
        {
        
    document.getElementById("txtHint").innerHTML=xmlhttp.responseText;
        }
      }
    xmlhttp.open("GET","response.php?q="+str,true);
    xmlhttp.send();
    }
    </script>
    </head>
    <body>

    <h3>Start typing a name in the input field below:</h3>
    <form action=""> 
    First name: <input type="text" id="txt1" onkeyup="showHint(this.value)" />
    </form>


    </body>
    </html> 
    Last edited by Inigoesdr; 11-13-2011 at 03:01 AM.

  • #2
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    Maybe the CSS style for your "hints" class is normally hidden or invisible?

    Show us your style sheet.

  • #3
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    it does not have any css for it yet

  • #4
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    Why do you need a <div> with the class called "hints"?
    Especially if you have no CSS stylesheet.

  • #5
    Super Moderator Inigoesdr's Avatar
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    You're missing the close quote on your first div line. You are getting an syntax error, and presumably display_errors is off so you get no output. Please remember to read the stickies for this forum. In particular the one about using [php][/php] tags when posting code.

  • Users who have thanked Inigoesdr for this post:

    kamkam (11-13-2011)

  • #6
    Master Coder
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    doh! I didn't see that ... good catch!

  • #7
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    Thanks


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