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pure_skill_2000

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Hi

 

This is a really simple php question I am sure but I am new to php and no matter what I do or where I look I cannt seem to make the code work, all I want is a simple search to allow a user to enter there forname and it be searched to return the results, below is the code I have which will not work

 

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml">

<head>

<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />

<title>Week 7 Ex 3</title>

</head>

 

<body>

<FORM NAME="searchphone" METHOD="post" ACTION="results.php">

Enter First Name: <input name="Array[Forname]" type="text" id="Array[Forname]"><BR>

<input type="submit" name="Submit" value="Submit">

 

</FORM>

 

</body>

</html>

 

results.php

 

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml">

<head>

<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />

<title>Untitled Document</title>

</head>

 

<body>

<?php

$Array["Forname"] =trim ($Array["Forname"]);

 

$Host="";

$User="";

$Password="";

$DBName="";

$TableName="";

 

$Link = mysql_connect($Host, $User, $Password);

 

$Query="SELECT * from $TableName Where Forname = $Array[Forname]";

 

 

$Result = mysql_db_query($DBName, $Query, $Link) or die(mysql_error());

 

while($row = mysql_fetch_array($result)){

 

print (" Name:$Row[surname], $Row[Forname] Phone: $Row[Telephone]<BR> ");

}

mysql_close ($Link);

?>

 

 

</body>

</html>

 

Of course in my code I have put in the connection details but I do not want to display them on the internet

 

Many thanks

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https://forums.phpfreaks.com/topic/82364-simple-search/
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$Query="SELECT * from $TableName Where Forname = $Array[Forname]";

 

Is the casing correct in that statement? MySQL is Case sensitive

 

allso, I would add ` around tablename and Forname

 

Example:

 

$Query="SELECT * from `$TableName` Where `Forname` = $Array[Forname]";

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https://forums.phpfreaks.com/topic/82364-simple-search/#findComment-418694
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