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Displaying PHP result within HTML?

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Displaying PHP result within HTML?

Apr 3rd, 2008

I've searched through the forum, but I couldn't find an answer. How do you display a result within HTML?

Such as:

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
$query = "SELECT * FROM customers WHERE job_number='$jobnumber'";
	$result = mysql_query($query);
	while ($row = mysql_fetch_array($result)){
		echo $row['first_name'];
		echo $row['last_name'];
$query = "SELECT * FROM customers WHERE job_number='$jobnumber'";
	$result = mysql_query($query);
	while ($row = mysql_fetch_array($result)){
		echo $row['first_name'];
		echo $row['last_name'];

Thanks in advance.

Last edited by peter_budo; Apr 5th, 2008 at 1:28 pm. Reason: [icode] is for one line of code, [code] is for multiple lines

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Re: Displaying PHP result within HTML?

Apr 3rd, 2008

Could you please explain more what you want to display?

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Re: Displaying PHP result within HTML?

Apr 3rd, 2008

I want to display the ['first_name'], ['last_name'], ['email'] fields. Here's the script I have used to call up a result from my database, and the result is the echo $row['first_name'] , etc. I'm just trying to figure out how to wrap the result within HTML.

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
if(isset($_POST['submit'])) {
	$jobnumber=$_POST['jobnumber'];
 
	//connect
 
	$query = "SELECT * FROM customers WHERE job_number='$jobnumber'";
	$result = mysql_query($query);
	while ($row = mysql_fetch_array($result)){
		echo $row['first_name'];
		echo $row['last_name'];
		echo $row['email'];
		//etc...
	}
}
?>
if(isset($_POST['submit'])) {
	$jobnumber=$_POST['jobnumber'];

	//connect

	$query = "SELECT * FROM customers WHERE job_number='$jobnumber'";
	$result = mysql_query($query);
	while ($row = mysql_fetch_array($result)){
		echo $row['first_name'];
		echo $row['last_name'];
		echo $row['email'];
		//etc...
	}
}
?>

Did I help clear it a little? Thanks for your help...

Last edited by peter_budo; Apr 5th, 2008 at 1:28 pm. Reason: [icode] is for one line of code, [code] is for multiple lines

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Re: Displaying PHP result within HTML?

Apr 3rd, 2008

there's no problem in your code that will display the results alright.Do you mean that you want to display the results in a specific area of an HTML page?

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Re: Displaying PHP result within HTML?

Apr 3rd, 2008

I just want to display the results within a table or a list.

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Re: Displaying PHP result within HTML?

Apr 3rd, 2008

just put the echo thing inside the table.

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
        $query = "SELECT * FROM customers WHERE job_number='$jobnumber'";
	$result = mysql_query($query);
	echo "<table>";
        while ($row = mysql_fetch_array($result)){
		echo "<tr>"
                echo "<td>$row['first_name']</td>";
		echo "<td>$row['last_name']</td>";
		echo "<td>$row['email']</td>";
                echo "</tr>";		
//etc...
	}
        echo "</table>";
        $query = "SELECT * FROM customers WHERE job_number='$jobnumber'";
	$result = mysql_query($query);
	echo "<table>";
        while ($row = mysql_fetch_array($result)){
		echo "<tr>"
                echo "<td>$row['first_name']</td>";
		echo "<td>$row['last_name']</td>";
		echo "<td>$row['email']</td>";
                echo "</tr>";		
//etc...
	}
        echo "</table>";

http://www.ryantetek.com/

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Re: Displaying PHP result within HTML?

Apr 4th, 2008

I seem to be getting this error: Parse error: syntax error, unexpected T_ECHO, expecting ',' or ';' in getjobnumber.php on line 30

Here's how I wrote it:

 Help with Code Tags 
 php Syntax (Toggle Plain Text) 
<?php
if(isset($_POST['submit'])) {
	$jobnumber=$_POST['jobnumber'];
 
	//connect
 
	$query = "SELECT * FROM customers WHERE job_number='$jobnumber'";
	$result = mysql_query($query);
	echo "<table>";
	while ($row = mysql_fetch_array($result))  {
		echo "<tr>"
		echo "<td>$row['first_name']</td>";
		echo "<td>$row['last_name'];</td>";
		echo "<td>$row['email'];</td>";
		echo "</tr>";
		//etc...
	}
		echo "</table>";
	}
?>
<?php
if(isset($_POST['submit'])) {
	$jobnumber=$_POST['jobnumber'];

	//connect

	$query = "SELECT * FROM customers WHERE job_number='$jobnumber'";
	$result = mysql_query($query);
	echo "<table>";
	while ($row = mysql_fetch_array($result))  {
		echo "<tr>"
		echo "<td>$row['first_name']</td>";
		echo "<td>$row['last_name'];</td>";
		echo "<td>$row['email'];</td>";
		echo "</tr>";
		//etc...
	}
		echo "</table>";
	}
?>