PHP not inserting data into MySQL

Question

I have looked at other posts and tried some of the suggestions but I still can't seem to populate a specific table in my database.

I will supply some blocks of my code to put things into perspective.

Here is my code for creating the table I want to populate with data:

    CREATE TABLE `Devices` (
      `dID` int(11) unsigned NOT NULL AUTO_INCREMENT,
      `deviceType` enum('BT','C1','AW') DEFAULT NULL COMMENT 'BT = Bluetooth, C1 = C1     Reader, AW = Acyclica Wifi',
      `deviceName` varchar(255) NOT NULL DEFAULT '',
      `deviceIP` varchar(15) NOT NULL DEFAULT '',
      `devicePort1` int(4) NOT NULL,
      `devicePort2` int(4) DEFAULT NULL,
      `deviceRealLatitude` float(10,6) NOT NULL,
      `deviceRealLongitude` float(10,6) NOT NULL,
      `deviceVirtualLatitude` float(10,6) DEFAULT NULL,
      `deviceVirtualLongitude` float(10,6) DEFAULT NULL,
      `deviceChanged` tinyint(1) DEFAULT NULL,
      `deviceStatus` tinyint(1) DEFAULT NULL,
      `deviceLastSeen` timestamp NULL DEFAULT NULL,
      `deviceBufferSize` int(4) DEFAULT NULL,
      `deviceSoftwareVersion` varchar(20) DEFAULT NULL,
      `deviceMacAddress` varchar(17) DEFAULT NULL,
      `deviceTest` tinyint(1) DEFAULT NULL,
      `deviceAPIKey` varchar(100) DEFAULT NULL COMMENT 'Only used with the Acyclia   Scanners',
      `deviceSerialID` int(6) DEFAULT NULL COMMENT 'Only used with the Acyclia  Scanners',
       PRIMARY KEY (`dID`)
      ) ENGINE=InnoDB AUTO_INCREMENT=13 DEFAULT CHARSET=latin1;

I have declared all the field variables ie

 $dID = 0;
 .....
 $deviceSerialID = 0;

I checked if each field is set ie

isset($_POST['dID']) == 1 ? ($dID = ($_POST['dID'])) : null;
.....
isset($_POST['deviceSerialID'])   == 1 ? ($deviceSerialID = ($_POST['deviceSerialID']))     : null;

Here is where I use INSERT INTO "table name" --> VALUES

$insert_query = "INSERT INTO Devices (`dID`, `deviceType`, `DeviceName`, `deviceIP`,
`devicePort1`, `devicePort2`, `deviceRealLatitude`, `deviceRealLongitude`,
`deviceVirtualLatitude`, `deviceVirtualLongitude`, `deviceChanged`, `deviceStatus`,
`deviceLastSeen`, `deviceBufferSize`, `deviceSoftwareVersion`, `deviceMacAddress`,
`deviceTest`, `deviceAPIKey`, `deviceSerialID`) VALUES ('".$dID."', '".$deviceType."',
'".$deviceName."', '".$deviceIP."', '".$devicePort1."', '".$devicePort2."', 
'".$deviceRealLatitude."', '".$deviceRealLongitude."', '".$deviceVirtualLatitude."', 
'".$deviceVirtualLongitude."', '".$deviceChanged."', '".$deviceStatus."', 
'".$deviceLastSeen."', '".$deviceBufferSize."', '".$deviceSoftwareVersion."', 
'".$deviceMacAddress."', '".$deviceTest."', '".$deviceAPIKey."', 
'".$deviceSerialID."')";

mysqli_query($conn,$insert_query);

Now, in my html form, I only want to input data for some of the fields:

<fieldset>
<form name = "form1" id = "form1" action = "" method = "post">
Device ID:         <input type="text" name="dID"        >               <br>

Device Type:              <select name="deviceType">                    
              <option value = "BT">BT</option>
              <option value = "C1">C1</option>
              <option value = "AW">AW</option>  
              </select>                                             <br>

Device Name:       <input type="text" name="deviceName" >               <br>
Device IP:         <input type="text" name="deviceIP"   >               <br>
Device Port 1:         <input type="text" name="devicePort1">               <br>
Device Port 2:         <input type="text" name="devicePort2">               <br>

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

I get no errors when I enter my data and hit submit. I realize that some of the things I have done in the code are not necessary such as the tick marks. Its all been a matter of trying to see if it would fix my problem. Any suggestions as to what I can do please?

UPDATE 6/20/14

It seems that trouble lies in my CREATE TABLE syntax. When I replace type 'int' or 'float' with 'varchar', for the fields (dID is an exception), it updates the table. Anyone familiar with posting an 'int' or 'float' value?

UPDATE 6/20/14 A couple hours later...

Guys, it seems I have discovered the problem. It was a small technical issue and I complicated the code to drown in further error.

The idea follows this:

I do not need to worry about a field being set at this point, as many fields will remain null. I only need to fill a few out using this method for each field:

ie

$deviceType                 = $_POST['deviceType'];

Also, the using an 'int' or 'float' requires a difference in how to implement the INSERT INTO ... VALUES clause. If leaving it null (which I am for now), I cannot do this:

$insert_query1 = "INSERT INTO Devices (deviceChanged) VALUES ('$deviceChanged');

This iteration was preventing the table from being filled with its appropriate values.

Instead, fill it with null:

$insert_query1 = "INSERT INTO Devices (deviceChanged) VALUES (NULL);

PHP now inserts data into MySQL. Thanks for the help!

Answer 1

You are trying to insert a string in a place of a float.

You should remove the ticks from the non varchar columents. example devicePort1 is declared and int but the insert statement is sending a value with a tick like this , '".$devicePort1."', instead of this, because MySQL expects an int u should send the , ".$devicePort1.", , ie with out the ticks.

Also because ID is AUTO_INCREMENT You should not to try to set value to it. To check if your statement is correct echo $query , then try to execute that from MySQL console or PHPMyAdmin.

A good practice is to use prepared statements or PDO. check this PHP: PDO - Manual or Prepared Statements Manual . Those links are from Php.net and they contain a very good explanation with examples;

Answer 2

isset return boolean value (true/false) not integer or something else ( http://www.php.net//manual/en/function.isset.php )

And another mistake is about how you use ternary comparision. See more about that here: http://davidwalsh.name/php-shorthand-if-else-ternary-operators

So instead of:

isset($_POST['deviceName']) == 1 ? ($deviceName = ($_POST['deviceName'])) : null;

you should write

$deviceName = isset($_POST['deviceName']) ? $_POST['deviceName'] : null;

Also please consider using some sort of loop to assign these variables (if they are at all needed) and not duplicate your code in huge amounts.