get a specfic user details from mysql database using php?

[jgkgopi]

hi

i had database with field of name,title,post,content

 

i want to fetch the post and content for a specific user from giving name of that user by form help me to get that

 

ps just give me idea to how to do that/

 

<form id="form1" name="form1" method="post" action="view.php">
  <label>Name
  <input type="text" name="textfield" />
  </label>
  <p>
    <label>
    	<input type="submit" name="Submit" value="Submit" />
    </label>
  </p>
</form>

 

 

 

[spiderwell]

"SELECT `post`, `content` from `table` where `name` = ' " . $_POST['textfield']  . ";";

[kney]

Use something like this:

 

$username = $_POST['textfield'];
$query = "SELECT * FROM table_name WHERE userName=$username;

while($result == mysql_fetch_array($query)) {
//display
echo $result['title];
echo $result['post'];
echo $result['content'];
}

[jgkgopi]

 

 

$username = $_POST['textfield'];
$query = "SELECT * FROM table_name WHERE userName=$username;
while($result == mysql_fetch_array($query)) {
//display
echo $result['title];
echo $result['post'];
echo $result['content'];
}

 

where i have to use this code pls in the form

[kney]

You can use this code above ur form code

 

if(isset($_POST['Submit'])){
$username = $_POST['textfield'];
$query = mysql_query("SELECT * FROM table_name WHERE userName=$username");

while($result == mysql_fetch_array($query)) {
//display
echo $result['title];
echo $result['post'];
echo $result['content'];
}

}

[jgkgopi]

I have used this code

But it is in the same page

 

post title,content,conclusion are the column name used in my mysql datbase

 

this page is also view.php in action also am using view.php

<?php
//include 
if(isset($_POST['Submit'])){
$username = $_POST['textfield'];
$query = mysql_query("SELECT * FROM final WHERE name=$username");

while($result == mysql_fetch_array($query)) {
//display
echo $result['post title'];
echo $result['content'];
echo $result['conclusion'];
}
}
?>
<form id="form1" name="form1" method="post" action="view.php">
  <label>Name
  <input type="text" name="textfield" />
  </label>
  <p>
    <label>
    <input type="submit" name="Submit" value="Submit" />
    </label>
  </p>
</form>

 

 

[wildteen88]

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

You should be using the assignment operator (=) there.

[kney]

ur right ..

lol, sorry

[jgkgopi]

have changed like this

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

 

still it is in the same page,

i want to get the user from me and show the details of that particular  user

 

[wildteen88]

Your query is constructed incorrectly

$query = mysql_query("SELECT * FROM final WHERE name=$username");

$username needs to be wrapped in single quotes.

[jgkgopi]

Form code

<form id="form1" name="form1" method="post" action="v.php">
  <label>Name
  <input type="text" name="textfield" />
  </label>
  <p>
    <label>
    <input type="submit" name="Submit" value="Submit" />
    </label>

  </p>
</form>

 

PHP code

<?php 
mysql_connect("localhost","root",""); 
mysql_select_db("lasttry"); 
$username = $_POST['textfield'];
  echo '</br>';
$query = mysql_query("SELECT * FROM final WHERE name=`$username` ");

while($result = mysql_fetch_array($query)) {
//display
echo $result['content'];
echo $result['conclusion'];
}
?> 

 

Error getting on line 9

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

Warning: mysql_fetch_array() expects parameter 1 to be resource, boolean given in D:\wamp\www\mobile\v.php on line 9

[spiderwell]

wrong quotes my friend, you have back ticks ```````````` around a value and you need single quotes ' ' ' ' ' ' ', backticks are used around column names or table names

"SELECT * FROM `final` WHERE `name`='$username'

then you should be ok

[jgkgopi]

thank you got it