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Karyna

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Posts posted by Karyna

    string not converted to date

    in PHP Coding Help

    Posted

    PLEASEEE no matter what  i do... datepicker always post the dates as string i do whatever ways know convert string to date but no way!! how convert string to date so i can sarch in a mysql table... tnx in advanced 

        <!--HTML-->

        <div class="control-group">
            <label class="control-label">from</label>
            <div class="controls input-append date form_date" data-date="" data-date-format='yyyy/mm/dd' data-link-format="yyyy/mm/dd">
            <input size="16" type="text" value="" id="from" name="from">
            <span class="add-on"><i class="icon-th"></i></span>
        </div>
            <input type="hidden" id="d" value="" /><br/>
        </div>
        <div class="control-group">
            <label class="control-label">tto</label>
            <div class="controls input-append date form_date" data-date="" data-date-format='yyyy/mm/dd' data-link-format="yyyy/mm/dd">
            <input size="16" type="text" value="" id="tto" name="tto">
            <span class="add-on"><i class="icon-th"></i></span>
        </div>
            <input type="hidden" id="h" value="" /><br/>
        </div>
        <div class="card-body">        
            <button type="submit" class="btn btn-secondary">search...</button>
        </div>

        <!--PHP-->

        $yfromx=$_POST['from']  ;
        $yuntilly=$_POST['untill']  ;
        $yfroma=str_replace(" ","/","$yfromx");
        $yuntillb=str_replace(" ","/","$yuntilly");

        $t1=gettype ( $yfromx ) ;
        $t2=gettype ( $yuntilly ) ;
        $yfrom=date('Y/m/d', '$yfroma');
        $yuntill=date('Y/m/d', '$yuntillb');
        $t3=gettype ( $yfrom ) ;
        $t4=gettype ( $yuntill ) ;
        echo $yfroma.' | '.$yuntillb.' | '.$t1.' | '. $t2.' | '.$t3.' | '. $t4;
        //SELECT username FROM oc_calendar WHERE start_date between '$from' AND '$to'
        $queryfebus = "SELECT distintc FROM glamm_pedidos WHERE fecha_folio between '$yfrom' AND '$yuntill';";
        //$queryfolb = "SELECT *M glamm_pedidos WHERE procesado='$yfoliox' or folio='$yfoliox' or nombre like '%$yfoliox%' or mail like '%$yfoliox%';";
        $resultfebus = mysqli_query($con,$queryfebus) or mysqli_error($con);
        blah 
        blah

            

            

    missing variable when uploading file

    in PHP Coding Help

    Posted · Edited by Karyna

    hi everybody... i cant explain why i have this error.... i do this upload files thing many times but now... i have 2 php files one is like this:

    *****************

    <form id="form1" name="form1" method="post" action="go_upload.php">
      <table width="900" border="0" cellspacing="0" cellpadding="0">
        <tr>
          <td><p>Imagen de Muestra:
            <input name="filename1" type="file" id="filename1" value="..."  size="60" />
          </p>

         bla

         bla

         bla

         <input type="submit" name="button" id="button" value="Upload File" />

    </form>

     

    the other file is like this

    go_upload.php

    ***********************

    $target = "models_bags/";
     $target = $target . basename( $_FILES['filename1']['name']) ;   <- line 55
     $ok=1;
     if(move_uploaded_file($_FILES['filename1']['tmp_name'], $target))   <- line 57

     {
     echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded";
     }
     else {
     echo "Sorry, there was a problem uploading your file.";
     }

     

     

    THE ERROR WHEN I RUN THE FILES IS:

    Notice: Undefined index: filename1 in C:\Mywebsites\bagsite\go_upload.php on line 55

    Notice: Undefined index: filename1 in C:\sitiosweb\imprebolsa\go_upload.php on line 57
    Sorry, there was a problem uploading your file.

     

    tnx for help me plssss

    Php Dont Send Meils

    in PHP Coding Help

    Posted

    i have a php form that recolect data form users and send some emails... i have no problems with that, but the part is drive me crazy its that as i show in the below example of my code only the last 2 mails are sent.. the first 2 not!!

    i been tried no quites, single quotes, double quotes... no use the first 2 mails are not sent.. only the last 2.. whats wrong?? php mail function dont accept variable names in the '$to' parameter??... the echo lines at the end i wroted so i sure the variables have the vaue wich its true... pls.. some one help in this!! :-\ :'(

     

    TNX

     

     

    .blah

    .blah

    .

    $meil2=$_POST['meil'];

    $jobmeil2=$_POST['jobmeil'];

    .blah

    .

     

    mail('$meil2',"VELEC 2012, Resgister succced!","{$datav}","From: info@velec.com");

    mail('$jobmeil2',"VELEC 2012, Register succed!","{$datav}","From: info@velec.com");

    mail("george.scott@stet.com","VELEC 2012, new customer has register !","$data_cust","From: info@velec.com");

    mail("support@stet.com","VELEC 2012, new customer has register !","$data_cust","From: info@velec.com");

     

    <?php

    echo $meil2;

    echo $jobmeil2;

    ?>

    Help Pls Find & Update

    in PHP Coding Help

    Posted

    im designin an "e-bay" like site and i need to do a page where administrator can activate or deactivate some pods.. so far im doing ok.. but have soe trouble with that page.. heres y code... the administrator writes de code of the post and click the button to search 4 it.. it fouds and then click a 2nd button to activate it means write an "1" in a field called "activo"... BUT IT DOESNT...

     

    <?php

     

    if(isset($_POST['button_a'])){

    $lacla = $_POST['anucod'];

    mysql_query("UPDATE anuncios SET activo=1 WHERE clave='$lacla'") or die(mysql_error());

    }

    ?>

     

    <?php

    if(isset($_POST['button'])){

    $lacla = $_POST['anucod'];

    $lacla2=trim('$lacla');

    mysql_select_db($database_clasif, $clasif);

    $cuerye="select * from anuncios where clave='$lacla'";

    $result = mysql_query($cuerye) or die(mysql_error());

    $row_anuncios = mysql_fetch_assoc($result);

    $totalRows_anuncios = mysql_num_rows($result);

    if (mysql_num_rows($result) > 0) {

    //User exists

    ?>

    <form id="form1" name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

    <p>

    <input type="text" name="anucod" id="anucod" />

    <input type="submit" name="button_a" id="button_a" value="ACTIVATE POST" />

    </p>

    <?php

    echo "<br>";

    echo "<br>";

    echo "<br>";

    echo "<br>";

    echo "<br>________________________________________________________";

    ?>

    </form>

    <?php

    } else {

    //User does not exist

    echo '0';

    }

    }

    ?>

    <form id="form1" name="form1" method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>">

    <p>

    <input type="text" name="anucod" id="anucod" />

    <input type="submit" name="button" id="button" value="enter post code..." />

    </p>

    </form>

    </body>

    Post Array Values

    in PHP Coding Help

    Posted

    hi all i have a problem i guess my code is right but i dont know what im doing wrong pls helpp

    i have one page lets say comiditis.php

     

    <form id="form1" name="form1" method="post" action="comoditi2.php">

    bla, bla, bla

    $totalRows_Recordset1 = mysql_num_rows($Recordset1);

     

    $comodi=$row_Recordset1['nm'];

    $arr = explode(",", $comodi);

    $i=0;

    foreach ($arr as &$value) {

    //$value = $value * 2;

    // echo $arr[$i]."<br />\n";

    $aidi=$arr[$i];

    ?>

    <label>

    <input name="comoditis" type="checkbox" id="<?php echo $i;?> " value="check" />

    <?php echo $aidi;

    ?> </label>

    <?php

    $i++;

    }

    ?>

    ***********************************************************

    comiditi2.php

    <?php

    // $ai=$_POST['comoditis']; <<<---- heres my problem!!!

    // echo $ai;

     

    ?>

     

     

    HOW I KNOW WHAT CHECKBOXES ARE CHECKED??? i cant even print the values of the array??? whatv im doing wrong?? tnx in advance

    drop menus in cascade

    in PHP Coding Help

    Posted

    hi guys im drivn nuts i was sure was easy but i do get it.. i have a table like this

    item              clasif            out

    -----------------------------------------

    apple          fruit              lbs

    orange        fruit              lbs

    pumpkins    vegetable    box

    banana        fruit              lbs

    pineapple    fruit              box

     

    i did 3 drop menus clasif  item  out  when the user select from 1st drop let say "fruit" the second drop must filter the database to all fruits and when select let say orange from 2nd drop the 3rd must contain just lbs... i try many ways but no one give me that filter!!!  plss will sombody helppp??

    chained dropdowns

    in PHP Coding Help

    Posted

    hi everyone

    i have a question i tought was easy but im having some problems i have 3 dropdowns menus the customer select one category from one of them.. let say "pets" the the other dropdown must be filled with pets (dogs, cats, parrots...) if the customer select dogs the 3rd dropdown must filled with (doberman, bulldog,chihuhaua, etc) i think u got the idea.. but im having trouble with the 2nd and 3rd dropdown... any ideas:::

     

    tnx in advncd

    filter database from words in an array

    in PHP Coding Help

    Posted

    TNX TO ALL but i finally get it

    the mistake was in the double quotes i this lines

     

    $word='"SELECT * FROM models WHERE descrip LIKE ' ."'";

                |____ this double quote must go!!!

    $srched=$_POST['words'];

    $pieces = explode(" ", $srched);

    $all=count($pieces)-1;

     

    for($i=0;$i<=$all;$i ++) {

        $word = $pieces[$i] ."'";

        if($i!=$all) {

          $word .= $word . ' OR descrip like ' . "'";

        } else {

          $word .= $word .'"' . ";";

                                      |__ this double quote mus t go to hell too!!!

        }

    }

     

    $colname=$word;

    mysql_select_db($database_zero1, $zero1);

    $query_srch = $colname;

    $buscar = mysql_query($query_srch, $zero1) or die(mysql_error());

    $row_srch = mysql_fetch_assoc($srch);

    $totalRows_srch = mysql_num_rows($srch);

     

     

    filter database from words in an array

    in PHP Coding Help

    Posted

    hi every1

    im driveing crazy with this... i have a database with a field with descriptions of products named "DESCRIP" in the main page i have a text field where user write the words to search for i wanrt make an

          "SELECT * FROM models where descrip like xxx OR descrip like xxx... bla bla bla"

     

    i wrote this code but send me an error... waht can be wrong???

     

    $word='"SELECT * FROM models WHERE descrip LIKE ' ."'";

    $srched=$_POST['words'];

    $pieces = explode(" ", $srched);

    $all=count($pieces)-1;

     

    for($i=0;$i<=$all;$i ++) {

        $word = $pieces[$i] ."'";

    if($i!=$all) {

    $word .= $word . ' OR descrip like ' . "'";

    } else {

    $word .= $word .'"' . ";";

    }

    }

     

    $colname=$word;

    mysql_select_db($database_zero1, $zero1);

    $query_srch = $colname;

    $buscar = mysql_query($query_srch, $zero1) or die(mysql_error());

    $row_srch = mysql_fetch_assoc($srch);

    $totalRows_srch = mysql_num_rows($srch);

     

    tnx in advcd

    php dont insert values in database mixed types

    in PHP Coding Help

    Posted

    im driving mad with this... i know my code is right but i must be missing something i have no syntax errors, the page doesn´tn display any error...i must save in a database each item buy it by customers.... but no values are stored i believe the insert to statement must be some wrong  the code i using is this

     

    num_rows was obtained by count the nmber of fields

    al fields are numeric except descrip and pictu are varchar

     

    for($i=0;$i<=count($estimate)-1;$i ++){

    if($estimate[$i] != NULL){

    $qty=$estimate[$i]['qty'];

    $descripp=$estimate[$i]['descrip2'];

    $ppictu=$estimate[$i]['pictu'];

    $pprice=$estimate[$i]['price'];

    $subt=$estimate[$i]['price']*$estimate[$i]['qty'];

     

    echo $i;  <-- it displays each number correct so the value of $estimate its OK!!

     

        mysql_query("INSERT INTO estimates (numb, name, mail, qty, descrip, pictu, unit, stotal, gtotal,item) VALUES

    ($num_rows, '$_POST[name]', '$_POST[mail]', $qty, $descripp, $ppictu,$pprice, $subt, $total, $i);");

     

    {

    ¨{

     

    also try this with same result...

     

    $cuery='"';

    $cuery.='INSERT INTO estimates (numb, name, mail, qty, descrip, pictu, unit, stotal, gtotal,item) VALUES (' ;

    $cuery.="'";

    $cuery.= '$num_rows';

    $cuery.="', ";

    $cuery.="'";

    $cuery.='$_POST[name]';

    $cuery.= "', '";

    $cuery.= '$_POST[mail]';

    $cuery.= "', '";

    $cuery.= '$qty';

    $cuery.="', ";

    $cuery.='"$descripp", ';

    $cuery.='"$ppictu", ';

    $cuery.= "'";

    $cuery.= '$pprice';

    $cuery.="', '";

    $cuery.='$subt';

    $cuery.="', ";

    $cuery.='$total';

    $cuery.="', ";

    $cuery.='$i';

    $cuery.=');';

    $cuery.='"';

    mysql_query("$cuery");

    php mail dont send images

    in PHP Coding Help

    Posted

    hi every one ;)

    im driving crazy maybe im missing sometihing i have a php code that give a shoping cart that my customers shop in my website... the final page look like this

    final.php displays:

     

        QTY                              DESCRIPTION                            PRICE

        ------|---------------------------------------|----------------------|-----------|

          3  |  professional noteboks            | (picture of item)  |  3.95    |

        ------|---------------------------------------|----------------------|-----------|

          2  | drawing paints                        | (picture of item) |  2.00  |

        ------|---------------------------------------|----------------------|-----------|

                                                                                  Total    |  5.95  |

                                      [sent to my mail]

     

    the button has the link to another php page (send.php) that has this code

    <?php

    $name=$_POST['name'];

    $mail=$_POST['mail'];

    $subject = "Send me this items";

    $message =  file_get_contents('http://www.mydomain.net/final.php') ; 

    $headers .= "Return-Path: mymail@hotmail.com\r\n";

    $headers .= 'MIME-Version: 1.0' . "\r\n";

    $headers .= 'Content-Type: text/html; charset=ISO-8859-1'."\r\n";

    $headers .= "From: mymail@domain.net\r\n";

     

    mail($mail, $subject, $message, $headers);

    ?>

     

    ALL ITS FINE!!! but only recieve an empty table NO IMAGES, NO INFO, NO LOGOS  :'( :'(

    am i missing something in the code??

     

    TNX A LOTT

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