PHP - Drop Down And Insert Into Mysql

Hello, I'm having a bit of a problem here, all help to this issues would be much appreciated

I am trying to use text boxes to insert numbers into the database based on what is inputed.
If I have a string, like this for example:
$variable = 09385493;

And I want to insert it into the database like this:
mysql_query("INSERT INTO integers(number)
VALUES ('$variable')");

When checking the integers table in my database, looking at the number field, the $variable that was inserted is outputted as
9385493

Notice the number zero was taken out of the front of the number. If the number is double 0's (009385493), both of those zero's would disappear, too.

Thanks

Hi guys

I have a registration form working fine, my database is as below:

userid
username
email
password
repeatpassword


I have added another column which is "name", users can update their profile once they have logged in so I have created updateprofile.php and when I login-->go to update profile and insert my name nothing adds to mysql name column

this is my code below:

<?php

include ("global.php");


//username session
$_SESSION['username']=='$username';
$username=$_SESSION['username'];

//welcome messaage
echo "Welcome, " .$_SESSION['username']."!<p>";

  if ($_POST['register'])
    {
     //get form data
     $name = addslashes(strip_tags($_POST['name']));

$update = mysql_query("INSERT INTO users (name) VALUES ('$_POST[name]') WHERE username='$username'");
   
}

?>

    <form action='updateprofile.php' method='POST'>
    Company Name:<br />
    <input type='text' name='name'><p />
 
    <input type='submit' name='register' value='Register'>
    </form>


can you please tell me where in this code is wrong? Im new in php so please excuse me if I have silly mistakes.

thanks in advance
 

I need help badly!

What I want to do is insert into database the value from the selected radio group buttons.. All of them. There are 10 radio groups total (they can be less, but not more).

Thanks!
Code: [Select]
<?php
   require_once('Connections/strana.php'); 
 mysql_select_db($database_strana, $strana);
 
 ?>
<link href="css/styles.css" rel="stylesheet" type="text/css" />

<table width="100%" height="100%" style="margin-left:auto;margin-right:auto;" border="0">
<tr>
<td align="center">
<form action="" method="post" enctype="multipart/form-data" name="form1">
<table>
<?php
$tema = mysql_query("SELECT * from prasanja where tip=2")or die(mysql_error());

function odgovor1($string)
{
$string1 = explode("/", $string);
echo $string1[0];
}
function odgovor2($string)
{
$string1 = explode("/", $string);
echo $string1[1];
}


while ($row=mysql_fetch_array($tema))
{
$id=$row['prasanje_id'];
$prasanje=$row['prasanje_tekst'];
$tekst=$row['odgovor'];
?>
    <tr>
    <td>
    </td>
    </tr>
    <tr>
    <td class="formaP">
 <?php echo $prasanje?>
   </td>
   </tr>
   <tr>
   <td class="formaO">
      <p>
        <label>
          <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor1($tekst) ?>"  />
          <?php odgovor1($tekst) ?></label>
        <br />
        <label>
          <input type="radio" name="Group<?php echo $id?>" value="<?php odgovor2($tekst) ?>" />
          <?php odgovor2($tekst) ?></label>
        <br />
      </p></td>
</tr>
<tr>
<td>
<br />

</td>
</tr>
<?php 




}
?>
</table>
<input align="left"type="submit" name="submit"
value="Внеси" >
</form>
</td>
</tr>
</table>
prasanje = question
tekst/odgovor = answer

The answer table:
id - primary
question_id - the questions ID whose answer is selected in the radio group
user_id - cookie takes care of this
answer - the value from radio group
date - automatic

I have this code: <?php
$con = mysql_connect("localhost","hhh","hhh");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("hhh", $con);

// --------------------
// Avatar insert check
// --------------------

session_start();

$name = $_POST[name];
$group = $_POST[group];
$age = $_POST[age];
$usernameid = $_SESSION[id];

$result = mysql_query("SELECT * FROM avatars WHERE name='$_POST[name]'");

$num = mysql_numrows($result);

if ($num == 0) {
mysql_query("INSERT INTO avatars (id, usernameid, name, group, age, xp)
VALUES ('', '$usernameid', '$name', '$group', '$age', '0')");

header( 'Location: me/' ) ;
}
else echo 'Sorry, please pick a new name';
?> And it does everything but put the data into the datebase. If I add a session befor and after '$request' they both run, but the sql doesn't. No error returns, if just redirects to the other page. Any help?

well this is truely embarrising...i have a insert statement which works within phpmyadmin but when using mysqli_query it returns a error.

INSERT INTO users (username, timestamp) VALUES ('test', UTC_TIMESTAMP())

  Unknown column 'timestamp' in 'field list'   i've been playing about with this for a few hours now ...tried changing the column name (timestamp), adding ` around column names as well as table name.   the column exists which is the strangest part, and ive even checked there is no space after the column name in the db.   whats going on please?  

I don't understand where the empty value is. I've substituted the variables for text and still have the same problem.

Code:
Code: [Select]
$sql = "INSERT INTO courses (course#, name, subject, semester, ap)VALUES('$courseNum', '$courseName', '$subject', '$semester', '$ap')";
Error:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1

hello can anyone help me how to insert GET variable into mysql   


<?php
$page_title = 'Personal Wellness';
include ('template/header.inc');
include_once('config.php');

$id = $_GET['id'];

if(isset($_POST['submit']))
//if submit was pressed
{
if(strlen($_POST['height'])<1) //if there was no height
{
print "You did not enter a height.";
}
else if(strlen($_POST['weight'])<1) //no weight
{
print "You did not enter a weight.";
}
else if(strlen($_POST['bodyfat'])<1) //no bodyfat
{
print "You did not enter a Body Fat Range";
}
else if (strlen ($_POST['bodywater'])<1) //no bodywater
{
print "You did not enter a Body Water Range";
}
else if( strlen($_POST['musclemass'])<1) //no musclemass
{
print "You did not enter a Muscle Mass";
}
else if (strlen ($_POST['physiqueratt'])<1) //no physiqueratt
{
print "You did not enter a Physique Ratings";
}
else if (strlen ($_POST['bonemass'])<1) //no bonemass
{
print "You did not enter a Bone Mass";
}
else if (strlen ($_POST['bmr'])<1) //no bmr
{
print "You did not enter a BMR";
}
else if (strlen ($_POST['basalmetabolic'])<1) //no basalmetabolic
{
print "You did not enter a Basal Metabolic Age";
}
else if (strlen ($_POST['visceralfat'])<1) //no visceralfat
{
print "You did not enter a Visceral Fat";
}
else if(strlen($_POST['registrationmonth'] && $_POST['registrationday'] && $_POST['registrationyear'])<1) // no date
{
print "You did not enter a date of birth";
}
else //all fields met
{
$id=$_GET['id'];
$height=$_POST['height'];
$weight=$_POST['weight'];
$bodyfat=$_POST['bodyfat'];
$bodywater=$_POST['bodywater'];
$musclemass=$_POST['musclemass'];
$physiqueratt=$_POST['physiqueratt'];
$bonemass=$_POST['bonemass'];
$bmr=$_POST['bmr'];
$basalmetabolic=$_POST['basalmetabolic'];
$visceralfat=$_POST['visceralfat'];
$date=$_POST['registrationyear'] . '-' . $_POST['registrationmonth'] . '-' . $_POST['registrationday'];

$insertadmin="INSERT into personalwelness (m_id,height,weight,body_fat,body_water,muscle_mas s,physique_ratt,bone_mass,bmr,basal_metabolic,visc eral_fat,evaluation_date) values ('$id','$height','$weight','$bodyfat','$bodywater','$mus clemass','$physiqueratt','$bonemass','$bmr','$basa lmetabolic','$visceralfat','$date')"; //registering admin in databae
echo $insertadmin;
$insertadmin2=mysql_query($insertadmin) or die("Could not insert admin");
print "Personal Wellness Successfully Submitted";
}
}

?>
<form method="post" class="form" action="<?php echo $_SERVER['PHP_SELF'];?>">
<fieldset><legend>Enter Personal Wellness Information in the form below:</legend>

<table width="80%" border="0">
<tr>
<td width="16%">Height(CM)</td>
<td width="2%">:</td>
<td width="82%"><label for="height"></label>
<input type="text" name="height" id="height" value="<?php if (isset($_POST['height']))
echo $_POST['height'];?>" /></td>
</tr>
<tr>
<td>Weight(KG)</td>
<td>:</td>
<td><label for="weight"></label>
<input type="text" name="weight" id="weight" value="<?php if (isset($_POST['weight']))
echo $_POST['weight'];?>" /></td>
</tr>
<tr>
<td >Body Fat Range</td>
<td>:</td>
<td><label for="body fat"></label>
<input type="text" name="bodyfat" id="bodyfat" value="<?php if (isset($_POST['bodyfat']))
echo $_POST['bodyfat'];?>" ></td>
</tr>
<tr>
<td>Body Water Range(%)</td>
<td>:</td>
<td><label for="bodywater"></label>
<input type="text" name="bodywater" id="bodywater" value="<?php if (isset($_POST['bodywater']))
echo $_POST['bodywater'];?>"/></td>
</tr>
<tr>
<td>Muscle Mass</td>
<td>:</td>
<td><label for="musclemass"></label>
<input type="text" name="musclemass" id="musclemass" value="<?php if (isset($_POST['musclemass']))
echo $_POST['musclemass'];?>"></td>
</tr>
<tr>
<td>Physique Ratings</td>
<td>:</td>
<td><label for="physiqueratt"></label>
<input type="text" name="physiqueratt" id="physiqueratt" value="<?php if (isset($_POST['physiqueratt']))
echo $_POST['physiqueratt'];?>"></td>
</tr>
<tr>
<td>Bone Mass</td>
<td>:</td>
<td><label for="bonemass"></label>
<input type="text" name="bonemass" id="bonemass" value="<?php if (isset($_POST['bonemass']))
echo $_POST['bonemass'];?>" /></td>
</tr>
<tr>
<td>BMR</td>
<td>:</td>
<td><label for="bmr"></label>
<input type="text" name="bmr" id="bmr" value="<?php if (isset($_POST['bmr']))
echo $_POST['bmr'];?>"/></td>
</tr>
<tr>
<td>Basal Metabolic Age</td>
<td>:</td>
<td><label for="basalmetabolic"></label>
<input type="text" name="basalmetabolic" id="basalmetabolic" value="<?php if (isset($_POST['basalmetabolic']))
echo $_POST['basalmetabolic'];?>"></td>
</tr>
<tr>
<td>Visceral Fat</td>
<td>:</td>
<td><label for="visceralfat"></label>
<input type="text" name="visceralfat" id="visceralfat" value="<?php if (isset($_POST['visceralfat']))
echo $_POST['visceralfat'];?>"></td>
</tr>
<tr>
<td>Evaluation Date</td>
<td>:</td>
<td> <?php echo date_picker("registration")?></td>
</tr>
</table>
</fieldset>
<div align="center"><input type="submit" name="submit" value="Submit" />

</div>
</form>
<?php
function date_picker($name, $startyear=NULL, $endyear=NULL)
{
if($startyear==NULL) $startyear = date("Y")-100;
if($endyear==NULL) $endyear=date("Y")+50;

$months=array('','January','February','March','Apr il','May',
'June','July','August', 'September','October','November','December');

// Month dropdown
$html="<select name=\"".$name."month\">";

for($i=1;$i<=12;$i++)
{
$html.="<option value='$i'>$months[$i]</option>";
}
$html.="</select> ";

// Day dropdown
$html.="<select name=\"".$name."day\">";
for($i=1;$i<=31;$i++)
{
$html.="<option $selected value='$i'>$i</option>";
}
$html.="</select> ";

// Year dropdown
$html.="<select name=\"".$name."year\">";

for($i=$startyear;$i<=$endyear;$i++)
{
$html.="<option value='$i'>$i</option>";
}
$html.="</select> ";

return $html;
}
?>

<?php
include ('template/footer.inc');
?>

How can i insert a image in a database and in my images folder using php?

I am trying to create a script that takes information from a form and puts in a database.
In the action page, I decided to post a URL that shows the user there story that they posted. This is where I ran into the problem. . I realized that for the optional fields I could not just use a seperate insert statement, because this creates a new row. So I desided to use update statments, but this STILL does not work, they values or simply not getting inserted.  Here is the code:
Code: [Select]
<?php

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

die('SPAM BOT!');
}
if (
!isset($_POST['title']) 
&& !isset($_POST['summary'])
&& !isset($_POST['story'])
&& !isset($_POST['rating'])
&& !isset($_POST['cat'])
)
{
die("<div id='impor'>You forgot to enter one(or more) of the following fields <br /> 
1. Title <br />
2. Summary <br />
3. Story<br />
 </div>
 ");

}


mysqlConnect();


//take data from form an\ put them in variable
$title_form = bb(mysql_real_escape_string($_POST['title'])); //required 
$summ_form = bb(mysql_real_escape_string($_POST['summary']));// required
$story_form = bb(mysql_real_escape_string($_POST['story'])); 
$cat_form = $_POST['cat'];
$rating_form = $_POST['rating'];
$username = $_SESSION['user'];
// Make the other var into a list of links




mysql_query("
INSERT INTO story_info (title, sum, story, user, cat, rating)
VALUES('$title_form','$summ_form', '$story_form,', '$username', '$cat_form','$rating_form')
");
if(isset($_POST['notes']))
{
$notes_form = mysql_real_escape_string($_POST['notes']);
$notes_final = bb($notes_form);
mysql_query("
UPDATE story_info
SET notes = '$notes_final'
WHERE story = '$story_form' AND user = '$username' AND sum = '$summ_form' AND title = '$title_form'
");
}
//put other in array. Use while loop to put link code. Then but it back into one non array variable
if(isset($_POST['u_id']))
{
$uid = mysql_real_escape_string($_POST['u_id']);
$uid_db = str_replace(' ','_', $uid);
$blerg = "
UPDATE story_info
    SET series_id = '$uid_db'
WHERE story = '$story_form' AND user = '$username' AND sum = '$summ_form' AND title = '$title_form'
";
mysql_query($blerg);

}
echo "<h1> Your Story Has Been Posted! Thanks for posting $username .   </h1>";
echo "Please review the post below <br />";
echo "<h2> $title_form </h2>";
echo "<strong> <h2> Summary: </h2> </strong> $summ_form";
echo "<h4> Story: </h4>";
echo "$story_form";
if(isset($notes))
{
echo "<h4> Author's Notes: </h4> ";
echo "$notes_final";


}
if (isset($uid_db))
{
echo '<h3> Unique Series ID </h3>';
echo '<p> Make sure to write down this! <br />' .$uid_db .'</p> ';
}

$db = mysql_query("
SELECT story_id
FROM story_info
WHERE story='$story_form' AND user='$username'
")or die(mysql_error());
$rows = mysql_fetch_assoc($db);

$id = $rows['story_id'];





echo "Catagory: $cat_form <br />
Rating: $rating_form  <br />
";
echo "<a href='?p=page&id=$id'> Click here to view your story! </a>'";
?>


Please help!

Hello Friends,

Now i m stuck with simple insert query .

here's the code

Code: [Select]
<?php 
include_once("conf.php");

$firstName=$_POST['fname'];
$lastName=$_POST['lname'];
$email=$_POST['email'];
$dob=$_POST['dob'];
$password=$_POST['pass'];

$fname= stripslashes($firstName);
$lname=stripslashes($lastName);


$mail= mysql_real_escape_string($email);
$password= mysql_real_escape_string($password);

mysql_select_db('site');
$statement="Insert into Accounts(Name,lastName,emailId,DOB,password) VALUES($fname,$lname,$mail,$dob,$password)";

$query=mysql_real_escape_string($statement);

mysql_query($query) or die("Cannot save data:</br> ".mysql_error());
echo "Data Saved Successfully";

?>

Now please explain me why i m getting the following error

Cannot save data:
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '@gmail.com,1988/12/20,password)' at line 1

Any help will be highly appreciated!

I've got

Code: [Select]
for ($i=1; $i<=5; $i++)
{
if(isset($_POST['partsusedqty'.$i]) && $_POST['partsusedqty'.$i] != "" && $_POST['partsusedqty'.$i] != "0.00") {
mysql_query("INSERT INTO partsused (ptnumber, partqty, partdesc, partprice)
VALUES ($ticket, '$partsusedqty'.$i, '$partsuseddesc'.$i, '$partsusedprice'.$i)")
or die(mysql_error()); }
}
I need to know the correct formatting to put these variable variables as values in the mysql query. With this particular code, I get the error "You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '.1, ''.1, ''.1)' at line 2" I've tried formatting this an endless number of ways, but I used this particular example because its the one I really thought should work. Everything I've tried that doesn't throw an error put the $partsusedqty in the partqty, partdesc, and partprice fields. Thanks for any help!

Hey guys for some reason this code is not working i can't see a problem myself could someone please have a look and point the issue out to me. what i mean by it not working is it won't insert into the database or show a mysql_error.

thanks in advance

Code: [Select]
$guestip = $_SERVER['REMOTE_ADDR'];
$time = date('G:i');
$date = date("y-m-d");

$query = mysql_query("SELECT * FROM IP_Address") or die(mysql_error());
while($row = mysql_fetch_assoc($query)){
   
    if($guestip != $row['ip']){
        //insert into db.
        mysql_query("INSERT INTO IP_Address(id, ip, date, time) VALUES(NULL,'$questip','$date','$time')") or die(mysql_error());
        echo "inserted in to database"; echo mysql_error();
    }else{
        // add hit count and update time and date.
        echo "already in db";
    }
}

guys, im having a problem here thats making me crazy

im making a system and i want to insert the current date in the mysql. I have the field called 'data' in the DB, but when i make the code to insert all the other fields, including the 'data', its work perfectly... unless that damn date!

$query = "INSERT INTO news (id, titulo, mensagem, data) VALUES (NULL, '$titulo', '$mensagem', 'date(\"d/m/Y\")')";

whats wrong with that?
its stores in DB as '0000-00-00'.
 

Array ( [data] => Array ( [0] => Array ( [latitude] => 22.934566 [longitude] => 79.08728 [type] => county [distance] => 44.328 [name] => Narsinghpur [number] => [postal_code] => [street] => [confidence] => 0.5 [region] => Madhya Pradesh [region_code] => MP [county] => Narsinghpur [locality] => [administrative_area] => [neighbourhood] => [country] => India [country_code] => IND [continent] => Asia [label] => Narsinghpur, India ) ) )

Please Help ! Thanks in Advance