php - Parse error: syntax error, unexpected 'text' (T_STRING), expecting ',' or ';'

I am having trouble getting this to work. I have been able to call data from a dropdown menu and place it into a table and have it actively update without reloading the page. I am now trying to get the database information that is called to appear within a text input field or another drop down menu.

So basically I have a drop down menu that will call up user information, I am trying to get that information that is called to appear within another form so I can update it. Here is the code I am working with;

table7.php

Select a person:
$con=mysqli_connect("localhost","user","password","database");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}


$result = mysqli_query($con,"SELECT * FROM users");

while($row = mysqli_fetch_array($result)) {
echo "" . $row['uname'] . "";
mysqli_close($con);
}
?>




Person info will be listed here.


 

getuser2.php

$q = intval($_GET['q']);
$con=mysqli_connect("localhost","user","password","database");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}


$sql="SELECT * FROM users WHERE id = '".$q."'";
$result1 = mysqli_query($con,$sql);



echo "




";

echo "";


while($row1 = mysqli_fetch_array($result1)) {
echo "";
echo "";
echo "";
echo "";
echo "";
}

echo "";


echo "
Username
E-Mail
Info 1
" . $row1['email'] . "
" . $row1['info1'] . "
";

mysqli_close($con);
?> 

Within the getuser2.php code if you strip out the form input section and replace it with a call for uname using the format directly below for email and info it will display the data called from the database in standard text format.

However, I am encountering this error:

Parse error: syntax error, unexpected 'text' (T_STRING), expecting ',' or ';' in C:\wamp\www\getuser2.php on line 25

Answer

The problem is the following:

echo "";

You have double quotes inside a double-quoted string. PHP doesn't know where the string ends.

An easy fix (since you're not using variables inside the string anyway) is to change the double quotes to single quotes:

echo '';