PHP - Insert Function Not Working

HI,i am using java script to create a add row function in the php .but when the first row data can insert into database ,the 2nd row data cannot insert into database ,can help me to check my coding? thanks a lot


Code: [Select]
<SCRIPT language="javascript">
        function addRow(tableID) {
 
            var table = document.getElementById(tableID);
 
            var rowCount = table.rows.length;
            var row = table.insertRow(rowCount);
 
            var cell1 = row.insertCell(0);
            var element1 = document.createElement("input");
            element1.type = "checkbox";
            cell1.appendChild(element1);
 
         
            var cell2 = row.insertCell(1);
            var element2 = document.createElement("input");
            element2.type = "text";
            cell2.appendChild(element2);

var cell3 = row.insertCell(2);
            var element3 = document.createElement("input");
            element3.type = "text";
            cell3.appendChild(element3);

var cell4 = row.insertCell(3);
            var element4 = document.createElement("input");
            element4.type = "text";
            cell4.appendChild(element4);

var cell5 = row.insertCell(4);
            var element5 = document.createElement("input");
            element5.type = "text";
            cell5.appendChild(element5);

 
        }
 
        function deleteRow(tableID) {
            try {
            var table = document.getElementById(tableID);
            var rowCount = table.rows.length;
 
            for(var i=0; i<rowCount; i++) {
                var row = table.rows[i];
                var chkbox = row.cells[0].childNodes[0];
                if(null != chkbox && true == chkbox.checked) {
                    table.deleteRow(i);
                    rowCount--;
                    i--;
                }
 
            }
            }
catch(e) {
                alert(e);
            }
        }
 
    </SCRIPT>


THE PHP CODE
Code: [Select]
<?php 
require_once ('../../../Connections/admin_db.php'); 

mysql_select_db("admin_db");

if ((isset($_POST["Submit"])) && ($_POST["Submit"] == "Submit")) {
$i=0;
foreach($_POST['abc'] as $value )
{
   
                 $abc           = $_POST['abc'][$i];
 $level         = $_POST['level'][$i];
     $level_desc    = $_POST['level_desc'][$i];
      $pc_desc       = $_POST['pc_desc'][$i];
 
 
//Insert Data into Instructor Profile Info
$q = "INSERT INTO plo_pc(p_name,plo_id,plo_criteria,plo_level,level_dec,plo_desc)
VALUES ('$list','$plo_id','$abc','$level','$level_desc','$pc_desc') " ; 
 
mysql_query($q) or die(mysql_error()) ;
$i=$i+1;
}
}
?>

HI All,
I would like create query when I add some one is look what the last number in field and add one ?

example :
you can see the img in attach
IDT is primary key + auto increment
Customer_id is number field
customer _name is name

how can create this login

IDT    Customer_ID    Customer_name
112           5                    bbbb
113           6                    ccccc
114           7                     eeeee

Also I need the inster function and is look the last field in customer_ID and Increase one ( +1)


Any help please/

THanks

Basically I need to input data into two tables.

I am running 3 different query's but only 2 of them work. The other one doesn't.

None working query:
mysql_query("INSERT INTO users(username, password, email, pin, key) VALUES('$username', '$password', '$email', '$key', '$pin')");

Working querys:

mysql_query("DELETE FROM beta_keys WHERE keys_new='$key'");

**and**

mysql_query("INSERT INTO beta_keys(keys_used) VALUES('$key')");

So any ideas why the top one doesn't work but the bottom two do?

<?php
error_reporting(E_ALL^E_NOTICE);
     $connect = mysqli_connect("");//removed
    $doc = $_GET["doctor"];
    $username = $_GET["username"];
    $sql = "SELECT fname, lname from newpatient where username = '$username'";
     $result = mysqli_query($connect, $sql);
     $value = mysqli_fetch_row($result);   
     $fname = $value[0];
    $lname = $value[1];
    $totalcost = $_GET["totalcost"];
    $reason1 = $_GET["reason1"];
    $reason2 = $_GET["reason2"];
    $reason3 = $_GET["reason3"];
    $reason4 = $_GET["reason4"];
    $reason5 = $_GET["reason5"];
    $reason6 = $_GET["reason6"];
    $reason7 = $_GET["reason7"];
    $reason8 = $_GET["reason8"];
    $date = $_GET["date"];
   
    $reasons = array($reason1,$reason2,$reason3,$reason4,$reason5,$reason6,$reason7,$reason8);
    rsort($reasons);
    $reason1 = $reasons[0];
    $reason2 = $reasons[1];
    $reason3 = $reasons[2];
    $reason4 = $reasons[3];
   
    if(isset($_REQUEST["yes"]))
     {
        $sql1 = "SELECT * FROM appointments where doctor_name = '$doc' and time = '$time'";
        $result1 = mysqli_query($connect, $sql1);
        $num_rows = mysqli_num_rows($result1);
        if($num_rows > 0)
        {
           echo "Appointment Time already chosen. Select another time.";
            echo "<script language = 'javascript'>document.location.href='make_appointment.php?doc=$doc&username=$username'</script>";         
        }
        else
        {             
             $sql2 = "INSERT INTO appointments (username, time, doctor_name, cost, reason1_for_visit, reason2_for_visit,reason3_for_visit,reason4_for_visit, fname, lname) values ('$username','$date','$doc',$totalcost,'$reason1','$reason2','$reason3','$reason4','$fname','$lname')";
           $result2 = mysqli_query($connect, $sql2);
          if($result2)
              echo "This worked.";
          else
              echo "Insert did not work.";
             //echo "<script language = 'javascript'>document.location.href='registered_login_page.php?username=$username'</script>";
      }      
    }
    mysqli_close($connect);
?>

I have no idea and there is no reason why this should not be working.

im simply trying to add three variables into a database, and only one works.

the other two do not work for any reason i can find.

can someone point out my error, if any?

code:
<?php
$date = date("Y-m-d"); 
$dbc = mysqli_connect('localhost', 'root', '', 'timer')
or die('Error connecting to DB');
$query = @"INSERT INTO sessions (date, user, sessiontime) VALUES ('$date', '$user', '$sessiontime')";
$user = @$_GET['user'];
$sessiontime = @$_GET['clock'];
if (@$_GET['addDB'] == "Session Complete")
{
mysqli_query($dbc, $query)
or die( '<br>Query string: ' . $query . '<br>Produced error: ' . mysqli_error($dbc) );
}
?>

Form:
Code: [Select]
<label for="user"><b><em>Your name:&nbsp;&nbsp;</b></em></label><br /><input type="text" name="user" value="Admin/User" />
<input id="clock" name="clock" type="text" value="00:00:0" readonly><br>
<input id="startstopbutton" type="button" value="S t a r t" onClick="startstop();" style="font-weight:bold"><br> 
    <input type="submit" name="addDB" value="Session Complete" />
See, all the variables match up!?

I dont get what im doing wrong?

my SQL Query wont execute on on following lines:

Code: [Select]
$result = mysql_query("INSERT INTO 'gallery' ('image', 'memberid', 'caption') VALUES ('$newFileName', '$member_id', '$caption')")
    or die (mysql_error());
i get the following error:

Code: [Select]
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 ''gallery' ('image', 'memberid', 'caption') VALUES ('gallery/9074849_1.jpg', '1',' at line 1
here is my full code:
Code: [Select]
<?php
require_once('connect.php');

$rand = mt_rand(1,9999999);
$rand2 = mt_rand(1,9999999);
$member_id = $_SESSION['SESS_MEMBER_ID'];
$caption = $_POST["caption"];

if(isset($_FILES['uploaded']['name']))
{
$allowed_filetypes = array('.jpg','.gif','.bmp','.png','.jpeg');
$max_filesize = 524288; // Maximum filesize in BYTES (currently 0.5MB) 
    $fileName = basename($_FILES['uploaded']['name']);
$errors = array();
$target = "gallery/";
$fileBaseName = substr($fileName, 0, strripos($fileName, '.'));

   // Get the extension from the filename.
   $ext = substr($fileName, strpos($fileName,'.'), strlen($fileName)-1); 
 
   //$newFileName = md5($fileBaseName) . $ext;
   $newFileName = $target . $rand . "_" . $member_id.$ext;
   
   // Check if filename already exists
   if(file_exists("gallery/" . $newFileName)) 
    {
     $errors[] = "The file you attempted to upload already exists, please try again.";
    }
   // Check if the filetype is allowed.
   if(!in_array($ext,$allowed_filetypes))
    {
         $errors[] = "The file you attempted to upload is not allowed.";
    }
    
// Now check the filesize.
   if(!filesize($_FILES['uploaded']['tmp_name']) > $max_filesize)
    {
         $errors[] = "The file you attempted to upload is too large.";
    }
  
    // Check if we can upload to the specified path.
   if(!is_writable($target))
    {
         $errors[] = "You cannot upload to the specified directory, please CHMOD it to 777.";
    }
 
    //Here we check that no validation errors have occured.
    if(count($errors)==0)
    {

//Try to upload it.
        if(!move_uploaded_file($_FILES['uploaded']['tmp_name'], $newFileName))
        {
            $errors[] = "Sorry, there was a problem uploading your file.";
        }
    }
    //Lets INSERT database information here
    //Here we check that no validation errors have occured.
    if(count($errors)==0)
    {
$result = mysql_query("INSERT INTO 'gallery' ('image', 'memberid', 'caption') VALUES ('$newFileName', '$member_id', '$caption')")
    or die (mysql_error()); 

{
            $errors[] = "SQL Error.";
        }
}
    //If no errors show confirmation message
    if(count($errors)==0)
    {
         echo "<div class='notification success png_bg'>
<a href='#' class='close'><img src='img/cross_grey_small.png' title='Close this notification' alt='close' /></a>
<div>
The file {$newFileName} has been uploaded<br>\n
</div>
</div>";


//echo "The file {$fileName} has been uploaded";
echo "<br>\n";
echo "<a href='gallery.php'>Go Back</a>\n";

   }
    else
    {
        //show error message
        echo "<div class='notification attention png_bg'>
<a href='#' class='close'><img src='img/cross_grey_small.png' title='Close this notification' alt='close' /></a>
<div>
Sorry your file was not uploaded due to the following errors:<br>\n
</div>
</div>";
//echo "Sorry your file was not uploaded due to the following errors:<br>\n";
        echo "<ul>\n";
        foreach($errors as $error)
        {
            echo "<li>{$error}</li>\n";
        }
        echo "</ul>\n";
echo "<br>\n";
echo "<a href='gallery.php'>Go Back</a>\n";
    }
    
}
else
{
    //Show the form
echo "Use the following form below to add a new image to your gallery;<br />\n";
    echo "<form enctype='multipart/form-data' action='' method='POST'>\n";
    echo "Please choose a file: <input name='uploaded' type='file' /><br />\n";
echo "Caption: <input name='caption' type='text' /><br />\n";
    echo "<input type='submit' value='Upload' />\n";
    echo "</form>\n";

//Echo Tests!
    echo "<br /><br />Random FileName: "; 
echo $rand;
echo "<br />";
echo "member ID: #";
echo $member_id;
}
?>
any help appreciated. its prob something simple.

my table has the following fields:

"gallery"
id (primary Key, AUTO_INC)
memberid (fetched from session)
image (will store image name including extension)
caption (from "caption" text field in form)

I have a mysql insert statement generated with php that is not populating the table.
I've echoed the statement and if I copy and paste into phpmyadmin it works fine.
The result of the mysql_query function is true.
I've emptied the table so there are no primary key conflicts.
I've put the statement in a try catch and it does not display a exception.
What else can I try?

Here's the statement
INSERT INTO `wp_term_relationships` (object_id, term_taxonomy_id, term_order) VALUES (1597,83,0)

Works absolute fine if I copy and paste into phpmyadmin. Does not populated the table if run through mysql_query

Can anyone help me out with this. I have been struggling on and off it for weeks.

Heres the description of the problem.

I insert into a MySQL database from a query that is performed using an array that is posted from a page. The incoming array contains selected users (unique id's) that are used to lookup the users telephone numbers in a query and then insert into a DB. This code works intermittently. It will insert the records fine and then on the next attempt it may not? There is not pattern to it working and failing. Does anyone have any idea why this is happening? I have tried a loop using the posted array data count and also a loop based on the count of records brought back from the query but to no avail, it is still intermittently working.

The '$stripped_message' data is just a text string.

Code: [Select]
<?php
$date = date('Y-m-d H:i:s');

for ($i=0;$i<count($_POST['recipient']);$i++)  // loop based on count of selected users

$uk_mob_number = "44".substr($row_selected_recipients['User_mobile'], 1); // take off the first digit and replace with 44
$unique = makeRandomstring();  // Create a unique string for each entry

$values.="('$unique','$date','$stripped_message','$uk_mob_number','sent','$date','$userid'),";
// usleep(50000); // Tried a delay to try to fix, did not work
} // End of loop

$values=substr($values,0,-1); // to remove last comma
$query="INSERT INTO `sms` (`sms_unique`, `sms_sent`, `sms_body`, `sms_to`, `sms_status`, `sms_db_entry_time`, `sms_user`) VALUES $values".";";
?>

This topic has been moved to JavaScript Help.

http://www.phpfreaks.com/forums/index.php?topic=308768.0

Is there any way to tell my php ajax file to run the update query if the data already exist and if not, then create the row in the database?  I have both the update and the insert functions created, but was just wondering if I could tell php which one to use without passing through a parameter.

Hello everybody, I can't seem to figure out why this insert code isn't working.  I'm trying to create a database of US zip codes.  I created this user interface (form) with nothing but a submit button to execute the insert query

<div id="right_content" class="">    


<h3> Insert Zips </h3>


<form action = "insertzip1.php" method = "post">

<input type = "submit" name = "submit" value = "submit"/>


</form>


</div> <!--closes right content-->



Well here is the insert query which is supposed to accomplish the task.  I have just included a tiny subsets of all the zipcodes (the insertzip1.php page which is the value of  the action attribute of the form). 


<php?



if (isset($_POST['submit'])) {


require ('config.php');




$query = "INSERT INTO zips (zip, lat, lon, city, state, county, z_type, xaxis, yaxis, zaxis, z_primary, worldregion, country, locationtext, location, population, housingunits, income, landarea, waterarea, decommisioned, militaryrestrictioncodes, decommisionedplace) VALUES


('00501', 40.81, -73.04, 'HOLTSVILLE', 'NY', 'SUFFOLK', 'UNIQUE', 0.22, -0.72, 0.65, 'Yes', 'NA', 'US', 'Holtsville, NY', 'NA-US-NY-HOLTSVILLE', '', 0, 0, '', '', 'No', '', ''),
('00501', 40.81, -73.04, 'I R S SERVICE CENTER', 'NY', 'SUFFOLK', 'UNIQUE', 0.22, -0.72, 0.65, 'No', 'NA', 'US', 'I R S Service Center, NY', 'NA-US-NY-I R S SERVICE CENTER', '', 0, 0, '', '', 'No', '', ''),
('00544', 40.81, -73.04, 'HOLTSVILLE', 'NY', 'SUFFOLK', 'UNIQUE', 0.22, -0.72, 0.65, 'Yes', 'NA', 'US', 'Holtsville, NY', 'NA-US-NY-HOLTSVILLE', '', 0, 0, '', '', 'No', '', ''),
('00544', 40.81, -73.04, 'IRS SERVICE CENTER', 'NY', 'SUFFOLK', 'UNIQUE', 0.22, -0.72, 0.65, 'No', 'NA', 'US', 'Irs Service Center, NY', 'NA-US-NY-IRS SERVICE CENTER', '', 0, 0, '', '', 'No', '', '')
";

$result = mysql_query($query);

header("Location: insertzipsuccess.php");

}else{ die ("Could not insert data because" . mysql_error());}

?>


The insertzipsuccess.php page is simply a page that prints out a success message if the query is successfully executed.  Well when I hit the submit button, I just get redirected to a blank insertzip1.php page
Can anyone show me what I'm not doing right here?

PS I already created the table with fields that correspond to all the fields I'm trying insert.

Sorry got it! my apologizes

Hello, guys. I am experiencing some problems with an INSERT statement in this page. It simply won't write to the database!
I added echo at the bottom to check my variables and they print the values just fine.
I checked the database, table and datafield names and everything is correct, plus I don't have any issues with the other 25 tables of my database.
I'm using XAMPP btw...

Any help would be appreciated!

Code: [Select]
<!DOCTYPE html>
<html>
<head>
  <meta content="text/html; charset=utf-8" http-equiv="content-type">
<title>Doctors</title>
<link rel="stylesheet" href="style.css" media="screen" />
</head>
<body >
<?php
session_start();
$inc_code=$_SESSION['incident'];
$doc_code=$_REQUEST['doctor_code'];
$con = mysql_connect("localhost","root","");
  if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
  $mdb = "registry_db";
  mysql_select_db($mdb, $con);
  mysql_query("SET NAMES 'utf8'", $con);
  
  
 
?>
 
<div id="myform">
<p>
   <h2>Doctor in charge</h2> 
  </p>
    <?php
    $sql="INSERT INTO doctors_per_incident(Incident_code, doctor_code) VALUES ('$inc_code', '$doc_code')";
    echo "1 record added"." ".$inc_code." ".$doc_code;
    mysql_close($con);
    ?>

</div>
</body>
</html>

i have been trying to insert date with when users post there comment but when i echo the date() with the comments..it just display 0000.00.00.00 just like that and when i checked my DB it was like that too..please what can i do.this is my code
Thankd in advance Code: [Select]
<?php
include"header.php";
if(isset($_POST['submit']))
{
$postdate=mktime(0,0,0,date("m"),date("d")+1,date("y"));
$comment=mysql_real_escape_string($_POST['comment']);
if($comment!=='')
{
$ins="INSERT INTO post(post_content,post_date)VALUES('$comment','$postdate')";
mysql_query($ins) or die(mysql_error());
}
else
{
echo"You can not post an empty page";
}
}

I am simply trying to use stripslashes for my mysqli insert statement, and errors are driving me nuts.. I've tried several variation and pattern with apostrophes and quotes to no avail. Should I even be using stripslashes to clean my data? Or is there a better function?

Notice: Use of undefined constant title - assumed 'title' in C:\wamp\www\php\simple_classifieds\add_posting.php on line 57

      $query = "INSERT INTO Postings (id, city_id, title, description) VALUES
('','$_POST[city]','" .  stripslashes($_POST[title])  . "','$_POST[description]')" or mysqli_error();