deslyxia

I gave this a shot but I get an error while ($row = mysql_fetch_assoc($result)){ echo "<tr> <td> <form id=".$row['ID']." name=".$row['ID']." method="post" action="p_select.php"> <input type="hidden" name="PID" value=".$row['ID']."> <input type="submit" value="Submit"> </form> </td> <td>".$row['FName']."</td> <td>".$row['LName']."</td> </tr>"; } the error I get is Parse error: syntax error, unexpected T_STRING, expecting ',' or ';'

I have a query that I have run. I am able to output the dataset into a simple table it updates on my page beautifully. In the following code what I want to be able to do is along with the other columns i output have a 4th column that contains a button on each row and when a user clicks a button on the page it selects that particular row. This is the code I have creating my table atm. echo "<table alingn=\"center\" width=\"400px;\">"; echo "<tr> <td>ID</td> <td>FirstName</td> <td>LastName</td> </tr>"; while ($row = mysql_fetch_assoc($result)){ echo "<tr> <td>".$row['ID']."</td> <td>".$row['FName']."</td> <td>".$row['LName']."</td> </tr>"; } echo "</table>"; Basically everytime it loops through I would want the row that shows the ID to be a button that would execute a php script but pass it that ID. I dont really want the ID shown on the page.

In my Table I have hundreds of LName. $q comes from a user input field on the main page. The purpose here is that every time a user types a letter in the field this query selects all rows from my table that contain the user input string.

I am getting the following code when executing a query: Parse error: syntax error, unexpected T_VARIABLE Below is the query it came from. //query db and build name array $qry="SELECT * FROM Patient WHERE LName LIKE "$q""; $result=mysql_query($qry); I know that the issue is coming from the double quotes around $q. If I switch them to single quotes I get no error... but I also get no results. I have looked up the difference between single and double quotes and I think i have it right. Single quotes use exactly what is in them ex. echo '$q'; would be $q Double quotes will evaluate the variable and use that ex. $q = 'php'; echo "$q"; would be php So this has me kind of lost as to what the issue is in my query.

I have a form on my web page. When a user clicks the submit button it executes an external php file. The job of this file is to connect to my db and query a table based on user inputs. All of this is working fine so far. What i need to figure out is how to take the result of that query (which could be multiple records from my table) back into a table the user can see on the original web page. The way i see it is i need to somehow pass the query result back to the main page and then ru na loop and dump everything into a table. My form looks like this <form id="searchForm" name="searchForm" method="post" action="psearch.php" autocomplete="off"> <table width="300" border="0" align="center" cellpadding="2" cellspacing="0"> <tr> <td width="112"><b>First Name</b></td> <td width="188"><input name="fname" type="text" class="textfield" id="fname" /></td> </tr> <tr> <td><b>Last Name</b></td> <td><input name="lname" type="text" class="textfield" id="lname" /></td> </tr> <tr> <td> </td> <td><input type="submit" name="Submit" value="Search" /></td> </tr> </table> </form> External PHP file <?php //Start session session_start(); //Include database connection details require_once('config.php'); //Array to store validation errors $errmsg_arr = array(); //Validation error flag $errflag = false; //Connect to mysql server $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } //Select database $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } //Function to sanitize values received from the form. Prevents SQL injection function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } //Sanitize the POST values $fname = clean($_POST['fname']); $lname = clean($_POST['lname']); //Query the Database $query = "select * from Patient where FName like '%$fname%' and LName like '%$lname%' "; $result = mysql_query($query) or die(mysql_error()); Code to build and fill my data table with the records. //Display Results echo "<table alingn=\"center\" width=\"400px;\">"; echo "<tr> <td>ID</td> <td>FirstName</td> <td>LastName</td> </tr>"; while ($row = mysql_fetch_assoc($result)){ echo "<tr> <td>".$row['ID']."</td> <td>".$row['FName']."</td> <td>".$row['LName']."</td> </tr>"; } echo "</table>"; ?> How do i get my query result from the external PHP file into the main html doc ... and build this table AFTER the form data is submitted?

psearch.php is just an external php file that i execute upon submit. It logs into the DB and performs a query. I somehow need to get my result set back onto the original page. How do i do this lol

Good Afternoon, The issue i have is as follows. I have a form on my webpage, WHen the user clicks the submit button it excecutes an external php file which querys my database. I need to get the results of that query back into my main web page in a table. In researching this It seems that i need to use AJAX and possibly JQuery to do this ... but i have never used either so im lost. Any help with this would be a great help as well as a great learning tool. <form id="searchForm" name="searchForm" method="post" action="psearch.php" autocomplete="off"> <table width="300" border="0" align="center" cellpadding="2" cellspacing="0"> <tr> <td width="112"><b>First Name</b></td> <td width="188"><input name="fname" type="text" class="textfield" id="fname" /></td> </tr> <tr> <td><b>Last Name</b></td> <td><input name="lname" type="text" class="textfield" id="lname" /></td> </tr> <tr> <td> </td> <td><input type="submit" name="Submit" value="Search" /></td> </tr> </table> </form> PhP File <?php //Start session session_start(); //Include database connection details require_once('config.php'); //Array to store validation errors $errmsg_arr = array(); //Validation error flag $errflag = false; //Connect to mysql server $link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD); if(!$link) { die('Failed to connect to server: ' . mysql_error()); } //Select database $db = mysql_select_db(DB_DATABASE); if(!$db) { die("Unable to select database"); } //Function to sanitize values received from the form. Prevents SQL injection function clean($str) { $str = @trim($str); if(get_magic_quotes_gpc()) { $str = stripslashes($str); } return mysql_real_escape_string($str); } //Sanitize the POST values $fname = clean($_POST['fname']); $lname = clean($_POST['lname']); //Query the Database $query = "select * from Patient where FName like '%$fname%' and LName like '%$lname%' "; $result = mysql_query($query) or die(mysql_error());