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date() problem

Discussion in 'PHP' started by moondancer, Sep 29, 2007.

  1. #1
    Hi, guys.
    I need some help for this piece of code:

    //define vars
    $date = $_POST['date'];
    $one = $_POST['one'];
    $two = $_POST['two'];
    $three = $_POST['three'];


    $d = date("D");

    $aa = time() + (1 * 24 * 60 * 60);
    $bb = time() + (2 * 24 * 60 * 60);
    $cc = time() + (3 * 24 * 60 * 60);
    $dd = time() + (4 * 24 * 60 * 60);
    $ee = time() + (5 * 24 * 60 * 60);
    $ff = time() + (6 * 24 * 60 * 60);
    $gg = time() - (6 * 24 * 60 * 60);
    $hh = time() - (5 * 24 * 60 * 60);
    $ii = time() - (4 * 24 * 60 * 60);
    $jj = time() - (3 * 24 * 60 * 60);
    $kk = time() - (2 * 24 * 60 * 60);
    $ll = time() - (1 * 24 * 60 * 60);

    if ($d=="Mon") {
    $varone = date('Y-m-d');
    $vartwo = date('Y-m-d', $aa);
    $varthree = date('Y-m-d', $bb);
    $varfour = date('Y-m-d', $cc);
    $varfive = date('Y-m-d', $dd);
    $varsix = date('Y-m-d', $ee);
    }

    else {
    if ($d=="Tue") {
    $varone = date('Y-m-d', $ll);
    $vartwo = date('Y-m-d');
    $varthree = date('Y-m-d', $aa);
    $varfour = date('Y-m-d', $bb);
    $varfive = date('Y-m-d', $cc);
    $varsix = date('Y-m-d', $dd);
    }

    else {
    if ($d=="Wed") {
    $varone = date('Y-m-d', $kk);
    $vartwo = date('Y-m-d', $ll);
    $varthree = date('Y-m-d');
    $varfour = date('Y-m-d', $aa);
    $varfive = date('Y-m-d', $bb);
    $varsix = date('Y-m-d', $cc);
    }

    else {
    if ($d=="Thu") {
    $varone = date('Y-m-d', $jj);
    $vartwo = date('Y-m-d', $kk);
    $varthree = date('Y-m-d', $ll);
    $varfour = date('Y-m-d');
    $varfive = date('Y-m-d', $aa);
    $varsix = date('Y-m-d', $bb);
    }

    else {
    if ($d=="Fri") {
    $varone = date('Y-m-d', $ii);
    $vartwo = date('Y-m-d', $jj);
    $varthree = date('Y-m-d', $kk);
    $varfour = date('Y-m-d', $ll);
    $varfive = date('Y-m-d');
    $varsix = date('Y-m-d', $aa);
    }

    else {
    if ($d=="Sat") {
    $varone = date('Y-m-d', $hh);
    $vartwo = date('Y-m-d', $ii);
    $varthree = date('Y-m-d', $jj);
    $varfour = date('Y-m-d', $kk);
    $varfive = date('Y-m-d', $ll);
    $varsix = date('Y-m-d');
    }

    else {
    if ($d=="Sun") {
    $varone = date('Y-m-d', $gg);
    $vartwo = date('Y-m-d', $hh);
    $varthree = date('Y-m-d', $ii);
    $varfour = date('Y-m-d', $jj);
    $varfive = date('Y-m-d', $kk);
    $varsix = date('Y-m-d', $ll);
    }
    }
    }
    }
    }
    }
    }


    Here comes the problem:

    $sql = "SELECT one, two, three, date FROM table WHERE date='$varone'";

    $result = mysql_query($sql) or die("Couldn't execute query");
    if(mysql_num_rows($result)) {
    while($row = mysql_fetch_row($result))
    {

    echo("<ul><li>$one</li><li>$two</li><li>$three</li></ul>");
    }

    } else {
    SEMrush
    echo("<p>No results for $varone.</p>");

    }


    If $varone exists the result I get is only HTML code:
    <ul><li></li><li></li><li></li></ul>

    if it doesn't:
    <p>No results for $varone.</p> (the expected one)

    I'm stuck. Please help!
     
    moondancer, Sep 29, 2007 IP
    SEMrush
  2. nico_swd

    nico_swd Prominent Member

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    #2
    Perhaps:
    
    echo("<ul><li>{$row['one']}</li><li>{$row['two']}</li><li>{$row['three']}</li></ul>");
    
    PHP:
    ??
     
    nico_swd, Sep 29, 2007 IP
  3. moondancer

    moondancer Peon

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    #3
    Thanks for the quick reply, but ... again: only the HTML code is parsed. :(
     
    moondancer, Sep 29, 2007 IP
  4. crazyryan

    crazyryan Well-Known Member

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    #4
    I can't see the purpose of this line:
    if(mysql_num_rows($result)) {

    Try..

    $sql = "SELECT one, two, three, date FROM table WHERE date='$varone'";

    $result = mysql_query($sql) or die("Couldn't execute query");
    while($row = mysql_fetch_row($result))
    {
    $one = $row['one'];
    $two = $row['two'];
    $three = $row['three'];

    echo '<ul><li>$one</li><li>$two</li><li>$three</li></ul>';
    }

    } else {

    echo '<p>No results for $varone.</p>';

    }
     
    crazyryan, Sep 29, 2007 IP
  5. nico_swd

    nico_swd Prominent Member

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    #5
    ^^ Note that variables won't be parsed between single quotes.
     
    nico_swd, Sep 29, 2007 IP
  6. moondancer

    moondancer Peon

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    #6
    :(

    I did COPY / PASTE of this code:

    $sql = "SELECT one, two, three, date FROM table WHERE date='$varone'";

    $result = mysql_query($sql) or die("Couldn't execute query");
    while($row = mysql_fetch_row($result))
    {
    $one = $row['one'];
    $two = $row['two'];
    $three = $row['three'];

    echo "<ul><li>$one</li><li>$two</li><li>$three</li></ul>";
    }


    and again - the result is only html


    -----
    if this helps - here is my SQL structure


    CREATE TABLE `table` (
    `id` int(11) NOT NULL auto_increment,
    `one` varchar(255) NOT NULL default '',
    `two` varchar(255) NOT NULL default '',
    `three` varchar(255) NOT NULL,
    `date` date NOT NULL,
    PRIMARY KEY (`id`),
    UNIQUE KEY `date` (`date`)
    )
     
    moondancer, Sep 29, 2007 IP
  7. moondancer

    moondancer Peon

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    #7
    Thank you, guys.

    I figure it out (at last)

    Here it is:

    $sql = "SELECT one, two, three, date FROM table WHERE date='$varone'";

    $result = mysql_query($sql) or die("Couldn't execute query");
    while($row = mysql_fetch_array($result))
    {
    $one = $row['one'];
    $two = $row['two'];
    $three = $row['three'];

    echo "<ul><li>$one</li><li>$two</li><li>$three</li></ul>";
     
    moondancer, Sep 29, 2007 IP