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Thread: query issue : print the second field of a my sql table

  1. #1
    Join Date
    Mar 2009
    Posts
    9

    query issue : print the second field of a my sql table

    Hello.
    I got this
    $sql = "SELECT * FROM notturna ORDER BY descrizione";
    $result = mysql_query ($sql);

    while ($row = mysql_fetch_row($result))
    {

    if ( $item == 0 ) { $_primafoto = $row[0]; }
    echo "photos[".$item."]=\"notturna/".$row[0].".jpg\";";


    $item=$item+1;
    }

    }
    ?>

    function arrow()
    {

    document.getElementById( "back2" ).style.display = "none";

    }

    function changePic(dir) {
    var image = document.images.photoslider,
    fwdBtn = document.getElementById('forward2'),
    backBtn = document.getElementById('back2'),
    n = photos.length-1;
    if (dir == "next") {
    which = (which < n) ? which + 1 : which;
    image.src = photos[which];
    backBtn.style.display = "inline";
    if (which == n) {
    fwdBtn.style.display = "none";
    }
    } else if (dir == "back") {
    which = (which > 0) ? which - 1 : which;
    image.src = photos[which];
    fwdBtn.style.display = "inline";
    if (which === 0) {
    backBtn.style.display = "none";
    }

    }
    return false;
    }


    </script>

    </head>

    <body OnLoad="arrow()">

    <div class="container2">
    <div id="logo" > <img title="logo" src="logo2.jpg"></div>

    <div class="menu">


    <a href="http://paolobergomifoto.altervista.org">Home </a>
    <a href="http://paolobergomifoto.altervista.org">Chi sono </a>
    <a href="http://paolobergomifoto.altervista.org/gallerie.html">Gallerie </a>
    <a href="http://paolobergomifoto.altervista.org/contatti.html">Contatti </a>
    <a href="http://paolobergomifoto.altervista.org">Credits </a>

    </div>

    <div id="backnotturna">

    <a href="#" onclick="return changePic('back');">
    <img id="back2" style="border:0px" src="indietro.jpg"></a>
    </div>
    <div class="centro">
    <div class="gruppofoto2"><a href="gallerymacro.php">Macro</a><a href="gallerypaesaggi.php">Paesaggi</a><a href="galleryritratti.php">Ritratti</a><a href="gallerybn.php">B&N</a><a href="gallerynotturna.php">Notturna</a><a href="galleryvarie.php">Varie</a><img src="notturna/<?php echo $_primafoto; ?>" name="photoslider">

    </div>
    </div>
    <div id="forward"><a href="#" onclick="return changePic('next');"> <img id="forward2" style="border:0px" src="avanti.jpg"></a>
    </div>
    <div class="inizio"><a href="#" onclick="which=1; changePic('back');return false" >Torna all'inizio della gallery</a>
    </div>
    <div id="footer">Created by Paolo Bergomi</div>
    </div>



    The table in db has also a second field that in the query is represented by the variable $row[1] . this is the description of the pictures.
    I have an issue cause i don't know how to do to show the content of this field using the query above.
    The pictures is correctrly showed in the gallery, and my target is to add the relative description.(as i wrote, is the second field in the table)
    any idea ?

    cheers
    paolo

  2. #2
    Join Date
    Aug 2006
    Location
    Michigan
    Posts
    1,046
    You could try this?

    PHP Code:
    if( $item == ) { 
       
    $_primafoto $row[0]; 
    }
    echo 
    "photos[".$item."]=\"notturna/".$row[0].".jpg\";";

    // Testing

    echo "Row[1]: " $row[1] . "<br />";
    echo 
    "Row[2]: " $row[2] . "<br />";
    echo 
    "Row[3]: " $row[3] . "<br />";
    echo 
    "Row[4]: " $row[4] . "<br />";
    // etc...

    // END

    $item=$item+1;


  3. #3
    Join Date
    Mar 2009
    Posts
    9
    hELLO phil
    thx really. But it does not work

    don't know why. i thought it wasgood as well but not indeed

    paolo

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