cannot insert row in table

[Pawan_Agarwal]

I can connect to database but when I try to insert row using submit button in PHP, it does run without any error and does nothing, pls help me out.......

 

if($_POST[query]!="")

{

$n1=$_POST['fname'];

$n2=$_POST['number'];

$n3=$_POST['email'].$_POST['domain1'].$_POST['domain2'];

$n4=$_POST['category'];

$n5=$_POST['query'];

$year=Date("Y");

$month=Date("m");

$day=Date("d");

$now=$year."-".$month."-".$day;

 

$sql=("INSERT INTO table_name (Name,Number,Email,Category,Query,Date) VALUES('$n1','$n2','$n3','$n4','$n5','$now')") or die(mysql_error());  

 

if (!mysql_query($sql,$connect))

  {

  die('Error: ' . mysql_error($connect));

  }else{echo $n1." ".$n2." ".$n3." ".$n4." ".$n5." ".$now;}

echo "1 record added";

} 

[mac_gyver]

how do you know your code is running? what output do you get?

[Pawan_Agarwal]

I am trying to insert the data from text box in my form to database, the whole code is running without any error.

 

The row cannot be inserted after execution of

 

 

$sql=("INSERT INTO table_name (Name,Number,Email,Category,Query,Date) VALUES('$n1','$n2','$n3','$n4','$n5','$now')") or die(mysql_error());  

 

if (!mysql_query($sql))

  {

  die('Error: ' . mysql_error($connect));

  }

echo "1 record added";

} 

[mac_gyver]

i know what you are trying to do. you need to tell us what you saw in front of you because that helps to pin down what is causing the problem. how do you know that the whole code is running?

[kicken]

Remove the or die() stuff after your query string. That is something you would use on the line where you run the query using mysql_query, not when you define the query text. Since you are checking mysql_query with an if statement though, you do not need it at all.

[Pawan_Agarwal]

<?php

date_default_timezone_set('Asia/Kolkata');

$username = "**********";

$password = "**********";

$hostname = "**********";

$database = "**********";

$connect = mysql_connect($hostname, $username, $password, $database) or die("Could not connect: ".mysql_error());

 

if ($connect) {

echo ("Connection is success <br>");////it executes

} else {

echo ("Connection is failed");

}

 

$selected = mysql_select_db($database,$connect) or die("Could not select database");

if ($selected) {

echo ("Connection is success to database<br><br>");////it executes

} else {

echo ("Connection is failed");

}

 

if($_POST['query']!="")

{

$n1=$_POST['fname'];

$n2=$_POST['number'];

$n3=$_POST['email'].$_POST['domain1'].$_POST['domain2'];

$n4=$_POST['category'];

$n5=$_POST['query'];

$year=Date("Y");

$month=Date("m");

$day=Date("d");

$now=$year."-".$month."-".$day;

echo $n1." ".$n2." ".$n3." ".$n4." ".$n5." ".$now; ///this is not working

 

$sql=("INSERT INTO table_name (Name,Number,Email,Category,Query,Date) VALUES('$n1','$n2','$n3','$n4','$n5','$now')") ///this is not working

 

if (!mysql_query($sql))

  {

  die('Error: ' . mysql_error($connect));

  }

echo "1 record added";

}

mysql_close($connect);

 

?>

 

 

 

I am trying this code and it is executing, but no row inserted after clicking on submit button...........thanks for helping me out......

[Barand]

Turn PHP error reporting on. You should be getting a syntax error with that code

[Pawan_Agarwal]

Thanks, I enabled the reporting error function.........

 

The syntax error it is showing is here in the mentioned line........

 

if(!$_POST['query']=="")

 

what might be done to correct this ??

[Barand]

you have no ; at the end of the statement on previous line

[Pawan_Agarwal]

 Undefined index: query in C:\xampp\htdocs\enquiry.php on line if($_POST['query']!=NULL)

[Pawan_Agarwal]

now there is no error, I am applying

 

 if(!empty($_POST['query']))

 

and this sorts error of syntax, however, I am still not able to insert data into database..........

[Pawan_Agarwal]

CODE:

$n1=$_POST['fname'];

$n2=$_POST['number'];

$n3=$_POST['email'].$_POST['domain1'].$_POST['domain2'];

 

Error:

Notice: Undefined index: fname in C:\xampp\htdocs\enquiry.php 

Notice: Undefined index: number in C:\xampp\htdocs\enquiry.php 

Notice: Undefined index: email in C:\xampp\htdocs\enquiry.php 

Notice: Undefined index: domain1 in C:\xampp\htdocs\enquiry.php 

Notice: Undefined index: domain2 in C:\xampp\htdocs\enquiry.php 

 

I am facing this error, what can be done here ???

[Csharp]

Are you sure the form you're using is sending that data like this? Would you mind pasting a copy of the HTML used for the form?

[Pawan_Agarwal]

<form action="enquiry.php" name="myform" method='post'> 

Name <input type="text" size="65" name="fname">

Cell Number <input type="text" size="65" name="number">

Email Address<input type="text" size="32" name="email">

 

<select style="align:center" name="domain1" size="1" style="width:50px">

 <option>@gmail</option>

 <option>@yahoo</option>

 <option>@hotmail</option>

 <option>@aol</option>

 <option>@inbox</option>

 <option>@fastmail</option>

 <option>@mail</option>

 <option>@lycos</option>

 <option>@care2</option>

 <option>@zenbe</option>

 <option>@zoho</option>

 <option>@aim</option>

 <option>@icloud</option>

</select>

 

<select align="center" name="domain2" size="1" style="width:80px">

 <option>.in</option>

 <option>.com</option>

 <option>.co.in</option>

</select>

 

Should I write $_REQUEST or $_POST???

 

CODE:

$n1=$_POST['fname'];

$n2=$_POST['number'];

$n3=$_POST['email'].$_POST['domain1'].$_POST['domain2'];

 

[Pawan_Agarwal]

some how, I solved it................now I am coding isset() instead of if(!$_POST['query']=="")