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ilikephp

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Posts posted by ilikephp

    request a list box

    in PHP Coding Help

    Posted

    Hello,

     

    I am using the send mail form, everything is sent to my mail except the country field, it is sending blank.

     

    $name = $_REQUEST['name'] ; 
$phone = $_REQUEST['phone'] ; 
$email = $_REQUEST['email'] ; 
$country = $_REQUEST['country'] ; 
$message = $_REQUEST['message'] ; 

$headers = "From: $email"; 
$sent = mail( $to, $subject, $name."\n".$phone."\n".$email."\n".$country."\n".$message, $headers);

     

    How could it be fixed plz?

     

    Thx

    add rel inside echo

    in PHP Coding Help

    Posted

    Hello,

     

    I need to add rel="..." in the first part of the script but it is not working,

     

    below is the script:

     

    echo " <a href='products/".$image[$abc]."' target='_blank'><img border=0 src='products/thumbs2/".$image[$abc]."' /></a>";

     

     

    Thank you,

    form validations

    in PHP Coding Help

    Posted

    Hello!

     

    When I fill my php form in internet explorer or firefox and click on submit, the spry validaitons will be displayed; but using google chrome or safari they are not displayed.

    I tried the form using safari but on localhost, they are displayed but over the internet the form will skip to the database and will not be validated.

     

    What could be the problem plz?

    echo on submit

    in PHP Coding Help

    Posted

    The form is too big, I just need to echo 'Fix your Errors please anywhere, could be next to the submit button'

     

    is this one correct?

    <?php
if($_POST['form1'] == 0){
    echo 'Check your errors';
} 
?>

    echo on submit

    in PHP Coding Help

    Posted

    Hi, I have a form, it is working fine and it validates using spry...

     

    When the user clicks the submit button and all the fields are filled, he will be redirected to the homepage page if not the required fields will be highlighted, I want to add a message next to the submit button or a popup that displays: "Plz check the errors"

     

    this is the code, I think I should add an echo somewhere ?!

     

    if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1")) {
  $insertSQL = sprintf(               
      GetSQLValueString($_POST['IP'], "text"));

.
.
.

  mysql_select_db($database_myform, $myform);
  $Result1 = mysql_query($insertSQL, $myform) or die(mysql_error());


  if (isset($_SERVER['QUERY_STRING'])) {
    $insertGoTo .= (strpos($insertGoTo, '?')) ? "&" : "?";
    $insertGoTo .= $_SERVER['QUERY_STRING'];
  }
header('Location: homepage.php');
}

?>

    checkbox group in php form

    in PHP Coding Help

    Posted

    I want them to be displayed in the same field with a comma to separate them.

     

    I put this code in the form:

    <input type="checkbox" name="father[]" value="answer 1" /> answer 1</label>  <br /><label>
<input type="checkbox" name="father[]" value="answer 2" /> answer 2 </label>  <br />

			<?php  $fatherValue = (isset($_POST['father']) ? implode(',', $_POST['father']) : '');  ?>

     

    and in the $insertSQL = sprintf("INSERT INTO ....

    GetSQLValueString($fatherValue, "text"),

     

     

    I dont know if I did something wrong but I am getting Null value in the Mysql database

    drop down list not displaying info online only on localhost

    in PHP Coding Help

    Posted

    Hi, I am using this code to display the info in mysql database to dropdown list, it is working good on the localhost, when I upload everything the form is still working fine, but the drop down list are not displaying anything!!

    What could be the problem?

     

    * I did upload all the tables also

     

    <?php
$query="SELECT ID, nationality FROM nationalities";
$result = mysql_query ($query);

echo "<select name=first_nationality_father value=''></option>";
echo "<option value=''>-- Choose one --</option>";
while($nt=mysql_fetch_array($result)){
echo "<option value=$nt[nationality]>$nt[nationality]</option>";
}
echo "</select>";
?>

    checkbox group in php form

    in PHP Coding Help

    Posted

    I tried to put this in my code:

     

    in the insert to database:

    GetSQLValueString($_POST['father'][1], $_POST['father'][2]),

     

    In the form:

    <label>
<input type="checkbox" name="father[]" value="answer 1" /> answer 1</label>  <br /><label>
<input type="checkbox" name="father[]" value="answer 2" /> answer 2 </label> <br />

     

     

    I got an error :S

    checkbox group in php form

    in PHP Coding Help

    Posted 

    <input type="checkbox" name="father" value="answer 1" />
answer 1</label>
                    <br />
                    <label>
<input type="checkbox" name="father" value="answer 2" />
                      answer 2</label>
                   

     

     

    php dropdown list box

    in PHP Coding Help

    Posted

    THANKS A LOT (Y)

     

    still have one issue:

     

    when the form is submitted successfully the success.php page will open

     

    $insertGoTo = "success.php";

     

     

    if someone tries to put this link www.mywebsite.php/success.php it will be opened, how can I restricted from being viewed unless the form is submitted?

     

     

    Thanks,

    php dropdown list box

    in PHP Coding Help

    Posted

    I am using the code below, it is working fine, but how can I add to the first line of the list an item without a value like: "Select please"

     

    <?php
$query="SELECT yearID,year FROM years";

$result = mysql_query ($query);
echo "<select name=education_level value=''>Education Level</option>";

while($nt=mysql_fetch_array($result)){
echo "<option value=$nt[yearID]>$nt[year]</option>";
}
echo "</select>";
?>

    php dropdown list box

    in PHP Coding Help

    Posted

    Hi,

     

    I am trying to call the data from Mysql but I am getting an empty drop down list, this is the code:

     

     

    mysql:

     

    create table years
(  yearID integer auto_increment,
   year varchar(30),
   primary key (yearID)
);

insert into years (yearID, year) values ('1', '2007-2008');
insert into years (yearID, year) values ('2', '2008-2009');
insert into years (yearID, year) values ('3', '2009-2010');
insert into years (yearID, year) values ('4', '2010-2011');
insert into years (yearID, year) values ('5', '2011-2012');
insert into years (yearID, year) values ('6', '2012-2013');

     

    PHP:

     

    <?php require_once('../Connections/connection.php'); ?>
<?php

$result = @mysql_query( "select yearID, year, from sss.years");

print "<p>Select a year:\n";
print "<select name=\"yearID\">\n";
while ($row = mysql_fetch_assoc($result)){
$yearID = $row[ 'yearID' ];
$year = $row[ 'year' ];
print "<option value=$yearID>$year\n";
}
print "</select>\n";
print "</p>\n";
?>

     

     

    Thank you!

    dependent drop down list

    in Javascript Help

    Posted

    this is the code:

     

    <html>
<body>
<script type='text/javascript'>
function update1(a)
{
if(a=='1' || a=='2')
{
document.getElementById("destination").innerHTML="<select name='combo2'><option>option A</option><option>Option B</option></select>";
}
}

</script>
<form name='form1'>
<table>
<p>First:</p>
<tr>
  <td><select name='combo1'  onChange='update1(this.value)'>
    <option value="" selected>Select</option>
    <option value="1">1</option>
    <option value="2">2</option>

  </select></td>
</tr>
<tr><td>
  <p><span id='destination'>
    <select name='combo2' disabled='disabled' >
      <option selected>Select combo 1 first</option>
    </select>
  </span></p>
  <p><script type='text/javascript'>
function update2(a)
{
if(a=='1' || a=='2')
{
document.getElementById("destination2").innerHTML="<select name='combo4'><option>option A</option><option>Option B</option></select>";
}
}

</script></p>
  <p> </p>
  <p>Second</p>
  <p> </p>
  <p>
    <select name='combo3'  onChange='update2(this.value)'>
      <option value="" selected>Select</option>
      <option value="1">1</option>
      <option value="2">2</option>
    </select>
  </p>
  <p>
    <span id='destination2'><select name='combo4' disabled='disabled' >
      <option selected>Select combo 1 first</option>
    </select></span>
  </p></td></tr>

</form>
</body>
</html>

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