PHP - Dymanic Drop Down Box / Insert

I'm missing something here. I have a form, and when the submit is pressed, the relevant post data inserts into table one, then I want the last insert id to insert along with other form data into a second table. The first table's still inserting fine, but I can't get that second one to do anything. It leapfrogs over the query and doesn't give an error. EDIT: I forgot to add an error: I get: 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 'usage, why VALUES ('14', '', '123', '','1234', '', '')' at line 1

query:INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company) VALUES ('test 14', 'asdfa', 'asdf', 'adf','asdf', '', '', '', '123', '', '')


Code: [Select]
if (empty($errors)) {

require_once ('dbconnectionfile.php');

$query = "INSERT INTO tbl_donar (donar_fname, donar_name, donar_address, donar_address2, donar_city, donar_state, donar_zip, donar_email, donar_phone, donar_fax, donar_company)
 VALUES ('$description12', '$sn', '$description4', '$cne','$description5', '$description6', '$description7', '$description8',
'$description9', '$description10', '$description11')";

$result = @mysql_query ($query);


if ($result) {

$who_donated=mysql_insert_id();

$query2 = "INSERT INTO tbl_donation (donor_id, donor_expyear, donor_cvv, donor_cardtype, donor_authorization, amount, usage, why)
 VALUES ('$who_donated', '$donate2', '$donate3', '$donate4','$donate5', '$donate6', '$donate7')";
$result2 = @mysql_query ($query2);

if ($result2) {echo "Info was added to both tables! yay!";}

echo "table one filled. Table two was not.";
echo $who_donated;

//header ("Location: http://www.twigzy.com/add_plant.php?var1=$plant_id");

exit();
} else {
echo 'system error. No donation added';

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 All,

I have 2 tables:

one
CarMake
- CarMakeID
- CarMakeDesc

two
CarModel
- CarModelID
- CarModelMake
- CarModelDesc

Depending on what the user selects in the first dropdown (carmake)
the possible selection in the second dropdown (model) needs to be limited to only the models from the selected carmake.
in the second table (Carmodel : the 'CarModelMake' = CarMakeID, to identify the make)

How do I limit the dropdown 'CarModel' based on the selected CarMake in the first dropdown.

link :    http://98.131.37.90/postCar.php


code :
-- -- --
Code: [Select]



<label>
            <select name="carmake" id="CarMake" class="validate[required]" style="width: 200px;">

          <option value="">Select CAR MAKE...</option>

            <?php while($obj_queryCarMake = mysql_fetch_object($result_queryCarMake)) { ?>
                <option value="<?php echo $obj_queryCarMake->CarMakeID;?>"
                <?php if($obj_queryCarMake->CarMakeID == $CarAdCarMake) { echo 'selected="selected"'; } ?> >
                <?php echo $obj_queryCarMake->CarMakeDesc;?></option>
            <?php } ?>

        </select>
</label>

        <label>
<select name="carmodel" id="CarModel" class="validate[required]" style="width: 200px;">

          <option value="">Select MODEL...</option>

            <?php while($obj_queryCarModel = mysql_fetch_object($result_queryCarModel)) { ?>
                <option value="<?php echo $obj_queryCarModel->CarModelID;?>"
                <?php if($obj_queryCarModel->CarmodelID == $CarAdCarModel) { echo 'selected="selected"'; } ?> >
                <?php echo $obj_queryCarModel->CarModelDesc;?></option>
            <?php } ?>

        </select>
</label>



I'm a novice.. and appreciates all the help !

I am creating a form that will allow the user to select the make of vehicle "FORD" for example. If that make of vehicle is selected among different makes of vehicles, then another box will appear, with all the models for that particular model "Fiesta" for example.

What type of code accomplishes this setup in my web page? I do not want to list 500 models in one drop down list, but just those for each make in the first drop down list.

Thanks much!

hi;


Code: [Select]
$giden = $_GET['sehir_part']+1;
$url = '31.php?sehir_part='.$giden.'';
//$erkek_top = 7
for($q=1;$q<=7;$q++)
{

//i want insert into to mysql in here ; but its doing only one inserting
so i want be 7 more insert

}
header( 'refresh:1;url=31.php?sehir_part='.$giden)and in every page loading i want inserting to mysql againg

pls helpme thanks

I'm trying to use PDO and get used to doing things this way. I've been away from php/mysql for a few years, so, I'm crusty.  I'm not getting any error messages back on this code, but the insert just doesn't happen.  My first guess is that I'm doing something wrong with the datetime now() function.  But, I may not have the PDO code right. I tried the script the old fashion way with mysql_query() and that worked.  So, it has to be something in this code. I believe my server is set up to do PDO as it shows:
PDO
PDO support   enabled
PDO drivers    mysql, sqlite

pdo_mysql
PDO Driver for MySQL, client library version   5.0.45

My php version is 5.2.14 and Mysql is 5.0.45.

Any help would be appreciated.

Code: [Select]
     $DBH = new PDO("mysql:host=$host;dbname=$dbname", $user, $pass);
     $DBH->setAttribute( PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION );

     $sql=$DBH->prepare("INSERT INTO assets(asset_name,date_added,short_desc) VALUES (:asset_name,NOW(),:short_desc)");

     $sql->bindParam(':asset_name',$asset_name);
     $sql->bindParam(':short_desc',$short_desc);

     $name=$_POST["input1"];
     $short_desc=$_POST["input2"];

     $DBH->exec();
     echo $name;

    echo "\nPDO::errorInfo():\n";
    print_r($DBH->errorInfo());
   }
   catch(PDOException $e) {
     echo "Syntax Error: ".$e->getMessage();
   }

Hello!Is this correct?

Code: [Select]
$sql="INSERT INTO pacienti (nume, prenume, cnp, varsta, sex, casa_asigurari, oras, strada, numar, judet, data_inregistrarii)
VALUES
('$_POST[nume]','$_POST[prenume]','$_POST[CNP]','$_POST[varsta]','$_POST[sex]','$_POST[casa]','$_POST[oras]','$_POST[strada]','$_POST[nr]','$_POST[judet]','$_POST[internare]')";

$sql="INSERT INTO diagnostic (nume)
VALUES
('$_POST[diagnostic]')";
and i mean..if i can do 2 insert in 2 different tables...or how can i do this?

Just a quick question I have a form, couple of drop down boxes, input fields etc.. at the end is the submit button. now my question is should my insert into database sqls go before the submit button inside the form?
out side the form? after the button inside the form?

If by any chance you could give me some advice on the best way to layout my sql that would be great. They are big and go into three tables, at the moment I have them as three seperate queries but I think they must be able to go into one. can someone please help me with the layout and syntax.

here is what the queries look like at the moment:


if ($IBselect=$_POST ['IBselect']);
{
//CUSTOMER
$enterCust="INSERT INTO customer(username, password, title, firstName, lastName, address, town, country, postCode, phone, email, dateCust) 
                        VALUES ('$_POST[username]', '$_POST[password]', '$_POST[title]', '$_POST[firstName]', '$_POST[lastName]', '$_POST[address]', '$_POST[town]', '$_POST[country]', '$_POST[postCode]', '$_POST[phone]', '$_POST[email]', 'DATE: Auto CURDATE()', CURDATE()";
$enterCust_query=mysql_query($enterCust)or die(mysql_error());

//CARD 
$enterCard="INSERT INTO card(cardNumber,  name, expDate, cardID) 
                        VALUES ('$_POST[cardNumber]','$_POST[name]', ' $_POST[expDate]', '$_POST[cardID]')";
$enterCard_query=mysql_query($enterCard);

//BOOKING
$RCenterBook=mysql_query("INSERT INTO booking (custNumber, cardID, rideName, seatNo1, seatNo2, price, price2, dateBook, ID_time_tbl) 
                VALUES ('$custNumber', '$_POST[cardID]', '$rideName', '$_SESSION[IBextract]', '$_SESSION[IBextract2]', '$IBendPrice1', '$IBendPrice2', 'DATE: Auto CURDATE()', CURDATE()',  '$_SESSION[IB_slot_Time]'");
$IBenterBooking_query=mysql_query($IBenterBook); 
        if (!mysql_query($enterCard, $enterCust, $IBenterBook))
        {
        die("Error:" .mysql_error());
        }
        echo "1 record added IB";
}


Thanks =)

I get the following message: Quote

name4Query failed: Unknown column 'late' in 'field list'

with this code.  What does it mean?
Code: [Select]
<?php
$apt=$_POST['search_term'];
$stat = mysql_connect("localhost","root","");
$stat = mysql_select_db("prerentdb");
$query = "SELECT name FROM payments WHERE late = 'L'";
$stat = @mysql_fetch_assoc(mysql_query($query));
echo $stat["name"];
$name=$_POST['name'];  
$apt=$_POST['apt']; 
$amtpaid=$_POST['amtpaid'];  
$rentdue=$_POST['rentdue'];  
$prevbal=$_POST['prevbal'];
$hudpay=$_POST['hudpay'];  
$tentpay=$_POST['tentpay']; 
$datepaid=$_POST['datepaid'];
$late=$_POST['late']; 
$comments=$_POST['comments'];  
$paidsum=$_POST['paidsum'];
$query = "
INSERT INTO payhist (name,apt,amtpaid,rentdue,prevbal,
hudpay,tentpay,datepaid,late,comments,paidsum)
VALUES('$name','$apt','$amtpaid','$rentdue','$prevbal',
'$hudpay','$tentpay','$datepaid','$late','$comments','$paidsum')";
$stat = mysql_query($query) or die('Query failed: ' . mysql_error());  
mysql_close();
echo "data inserted<br /><br />"; 
?>

hey developers,   i have problem to get last insert id. btw.. i try to put that code on my mvc creation.   on register_model page i write like this:  public function numbering(){
        $sth = $this->db->prepare('SELECT * FROM member_data');
        $result = $this->db->lastInsertId();
        return $result;
    }
  and for the controllers page, i write the code like this:   $c = $this->model->numbering();
         foreach ($c as $key => $value) {
$data = $result('id');             
}          
        
$abc['member_id'] = $data;     but i didn't get the results,.. can you help me?

Dear Sir/Madame

I am trying to insert comments into a table called comments using echo ID in the input form but so far i can insert the comments with the post id but unable re-comment again kindly help me what do i do? 

This is the input form:

 <div>
        <form action="ehscomment.php" method="post">
        <input type="hidden" name="id" value="<?php echo $id ?>"> 
            <div>
                <label>Add comment</label>
    <div>
        <textarea rows="6" cols="110" name="comment" placeholder="comment"></textarea>
    
        </div>
        </div>
            <input type="submit" name="postcomment" value="comment"></form>
</div> 

And this is ehscomment.php

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "registration";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
   die("Connection failed: " . $conn->connect_error);
} 
if(isset($_POST['postcomment'])){
       $id = $_POST['id'];
       $comment = $_POST['comment'];
       

$sql = "INSERT INTO comments (id,comment)
VALUES ('$id','$comment')";

if ($conn->query($sql) === TRUE) {
   echo "New record created successfully";
} else {
   echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();}
?>

Error message: Error: INSERT INTO comments (id,comment) VALUES ('242','asddd')
Duplicate entry '242' for key 'PRIMARY'

I am working with a insert query...

Code: [Select]
$query="INSERT INTO user_info_more (website) VALUES ('".$website."') WHERE id='$_SESSION[id]'";
mysql_query($query) or die ('Unable to register an account with following details 1');

but the output is " Unable to register an account with following details 1"

Basically i have a row with 6 cloumns in mysql database in which only the id column and name column is filled...
Now i want to insert $website in the "website" column of the database...

Neither i can update as the website column initially is null...

I do not know what to do....
A lot of cnfusion

Not being able to upload images in the directory with a path name in the database?

1) This is my code for insert into the database and directory:

<?php


$host="localhost";
$username="root";
$pass="";
$db="registration";

$conn=mysqli_connect($host,$username,$pass,$db);
if(!$conn){
	die("Database connection error");
}

// insert query for register page
if(isset($_POST['ronel'])){
     $images = $_FILES['file']['name'];
  $target_dir = "uploads/";
  $target_file = $target_dir . basename($_FILES["file"]["name"]);

    // Select file type
  $imageFileType = strtolower(pathinfo($target_file,PATHINFO_EXTENSION));

  // Valid file extensions
  $extensions_arr = array("jpg","jpeg","png","gif","pdf");

  // Check extension
  if( in_array($imageFileType,$extensions_arr) )
  {

$details=$_POST['details'];
$location=$_POST['location'];
$checkbox=$_POST['checkbox'];
$injured=$_POST['injured'];    
$agegender=$_POST['agegender'];   
$contact=$_POST['contact'];
$empid=$_POST['empid'];
$dept=$_POST['dept'];    
$organization=$_POST['organization'];     
$summary=$_POST['summary']; 
$name=$_POST['name'];
$outcome=$_POST['outcome'];    
$cause=$_POST['cause'];     
$action=$_POST['action'];
$reportedname=$_POST['reportedname'];
$position=$_POST['position'];    
$organisation=$_POST['organisation'];
$reportedcontact=$_POST['reportedcontact'];
$reporteddept=$_POST['reporteddept'];    
$status="Pending";
$comment=$_POST['comment'];
	
	
	 $query="INSERT INTO `proposals` (`details`,`location`,`date`,`time`,`checkbox`,`injured`,`agegender`,`contact`,`empid`,`dept`  ,`organization`,`summary`,`image`,`outcome`,`cause`,`action`,`reportedname`,`position`,`organisation`,`reportedcontact`,`reporteddept`,`status`,`comment`) VALUES ('$details','$location', current_timestamp(),current_timestamp(),'$checkbox','$injured','$agegender','$contact','$empid','$dept' ,'$organization','$summary','$name','$outcome','$cause','$action','$reportedname','$position','$organisation','$reportedcontact','$reporteddept','$status','$comment')";
	
	$res=mysqli_query($conn,$query);
	if($res){
		$_SESSION['success']="Not Inserted successfully!";
		header('Location:');
	}else{
		echo "<script>alert('Proposal not applied!');</script>";
	}
      // Upload file
     move_uploaded_file($_FILES['file']['tmp_name'],$target_dir.$image);
	}
}
date_default_timezone_set("Asia/Kolkata");
?>

2) Here is the input file:

<form class="form-horizontal" method="post" action="" enctype="multipart/form-data">
            <input type="hidden" name="ronel" value="">
  <div class="form-group">
  <label style="position:absolute; left:63%; top:425px;" for="inputEmail" class="col-lg-3"><b>Upload Images Here :</b></label><br><br>
  <div class="col-lg-9">
  <input style="position:absolute; left:78%; top:420px;" type="file" name="file" enctype="multipart/form-data" class="form-control" name="incident_reference" onchange="document.getElementById('inc_ref').src = window.URL.createObjectURL(this.files[0]); document.getElementById('inc_ref').className +='_active'; document.getElementById('inc_ref_span').className += '_hidden'">
  </div><iframe id="inc_ref" class="form-group" width="220px" height="130px" style="position:absolute; left:78%; top:32%;"></iframe></div>
</form>

3) error code:

Notice: Undefined index: name in /opt/lampp/htdocs/create-nearmiss.php on line 57
ONGC TRIPUR

 

Edited March 28, 2020 by Ronel
change of var

Okay,
I'm hoping one of you can help me.
I have a mysql database that I have configured through phpmyadmin.
I have an android app that simply makes and sends a mysql query
I can get it to successfully return values when using Select statements but when I use INSERT INTO, it returns " Error Query is invalid" BUT BUT BUT, when I use the same string and enter it through the sql tab in myphpadmin it works fine !

So here is the string ( the semicolons at the end of each field name are so I can use something common to split the string up when the data arrives back on the phone)
Code: [Select]
randomkey||||||INSERT INTO table4  (`geolat;` , `geolong;` , `mode;` , `destgeolat;` , `destgeolong;` , `cellphone;` , `email;` , `carrego;` , `colour;` ,  `rating;` , `comment;`)
VALUES (0.0,0.0,'driver' ,-43.54779,172.62472, , '' ,'' , 'text' , 'ratingleftblank' , 'commentblank' )
the index4.php script is as follows
Code: [Select]
?php
/*
 * Written By:
 * James
 */


/************************************CONFIG****************************************/
//DATABSE DETAILS//
$DB_ADDRESS="mysql1.openhost.net.nz";
$DB_USER="bling44";
$DB_PASS="sadlyinept";
$DB_NAME="bling44";

//SETTINGS//
//This code is something you set in the APP so random people cant use it.
$SQLKEY="randomkey";
/************************************CONFIG****************************************/

//these are just in case setting headers forcing it to always expire and the content type to JSON
header('Cache-Control: no-cache, must-revalidate');
header('Content-type: application/json');


if(isset($_POST['tag'])){ //checks ifthe tag post is there
$tag=$_POST['tag'];
$data=explode("||||||",$tag); //split the SQL statement from the SQLKEY
if($data[0]==$SQLKEY){ ///validate the SQL key
$query=$data[1];
$link = mysql_connect($DB_ADDRESS,$DB_USER,$DB_PASS); //connect ot the MYSQL database
mysql_select_db($DB_NAME,$link); //connect to the right DB
if($link){
$result=mysql_query($query); //runs the posted query (NO PROTECTION FROM INJECTION HERE)
if($result){
if (strlen(stristr($query,"SELECT"))>0) { //tests if its a select statemnet
$outputdata=array();
while ($row = mysql_fetch_assoc($result)){
$outputdata[]=$row; //formats the result set to a valid array
}
echo json_encode(array("VALUE",$tag,array_merge($outputdata))); //sends out a JSON result with merged output data
} else {
echo json_encode(array("VALUE",$tag,array_merge(array(array("AFFECTED_ROWS ".mysql_affected_rows($link)))))); //if the query is anything but a SELECT it will return the array event count
}
} else echo json_encode(array("VALUE",$tag,array_merge(array(array("ERROR QUERY IS INVALID"))))); //errors if the query is bad
mysql_close($link); //close the DB
} else echo json_encode(array("VALUE",$tag,array_merge(array(array("ERROR Database Connection Failed"))))); //reports a DB connection failure

} else {
echo json_encode(array("VALUE",$tag,array_merge(array(array("ERROR BAD CODE SUPPLIED"))))); //reports if the code is bad
}
}
?>
So to reiterate.
I can search the DB but can't INSERT INTO, unless I go through the myphpadmin interface.
Any ideas are very much appreciated

i need the insert code