Member Avatar for neethub3@gmail

code to link database selected through dropdownlist with corresponding database in mysql.Plz reply as soon as possible.
Dropdownlist will list all database names inserted in a particular table

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  • 4 Replies
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  • 11 Hours Discussion Span
  • Latest Post Latest Post by saiprem

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Member Avatar for colweb

What have you written so far, where does it go wrong. Please post the problem code so we all can have a look at it.

Member Avatar for neethub3@gmail 
//php code

<?
mysql_connect("localhost", "root", "123") or die(mysql_error());

mysql_select_db("jai") or die(mysql_error());

$sql="SELECT distinct db_name FROM dbsname";
$result=mysql_query($sql);

$options="";

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

    $db_name=$row["db_name"];
    $options.="<OPTION VALUE=\"$db_name\">".$db_name;
}
?>

<html>
   
<body background="lgrey013.jpg" bgcolor="#FAEBD7">
        <label> Databases Available </label>
      <SELECT NAME=db_name>

          <OPTION VALUE=0>Choose
<?=$options?>
</SELECT>

    </body>

</html>

iam only up to these .i dont have any idea regarding how to choose the selected database name in mysql and list the tables already existing.Plz help me

Edited by nav33n because: Please use [code][/code] tags to wrap your code for easy readability.
Member Avatar for colweb

Did re-write your code. The following should work but you may have to change some things in it.

<?php
// the html header
echo "<html>
<body background=\"lgrey013.jpg\" bgcolor=\"#FAEBD7\">";

// connect with database
mysql_connect("localhost", "root", "123") or die(mysql_error());
mysql_select_db("jai") or die(mysql_error());

// do the query
$sql="SELECT distinct db_name FROM dbsname";
$result=mysql_query($sql);

// create a form with select box
echo "form method=\"post\" action=\"whatever.php\">
<select name=\"db_name\">";

// The while loop is looping through the database rows,
// the for loop is looping through the current database row.
while ($row = mysql_fetch_assoc ($result)) {
	foreach ($row as $var => $value) {
		echo "option value=\"".$var."\">".$value."</option>";
	}
}
echo "</select>
<input type=\"submit\" value=\"submit-form\"/>";

// the html footer
echo "</body>
</html>";
?>
Edited by colweb because: n/a
Member Avatar for saiprem

Try like this, it will works for you

<?php
mysql_connect("localhost", "root", "123") or die(mysql_error());
mysql_select_db("jai") or die(mysql_error());

$sql = "SELECT distinct db_name FROM dbsname";
$result = mysql_query($sql);

$options="";

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

    $db_name = $row["db_name"];
    $options.="<OPTION VALUE=\"$db_name\">".$db_name."</OPTION><br />";
}
if(isset($_POST['db_name']) && $_POST['db_name']!='')
{
	//perform your other operations based on the selected distinct value
}
?>
<html>
<body background="lgrey013.jpg" bgcolor="#FAEBD7">
<form method="post" name="frmName">
      <SELECT NAME='db_name' onchange="javascript:this.form.submit()">
         <OPTION VALUE=0>Choose</OPTION>
			<?php echo $options; ?>
		</SELECT>
</OPTION>
</body>
</html>
//php code

<?
mysql_connect("localhost", "root", "123") or die(mysql_error());

mysql_select_db("jai") or die(mysql_error());

$sql="SELECT distinct db_name FROM dbsname";
$result=mysql_query($sql);

$options="";

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

    $db_name=$row["db_name"];
    $options.="<OPTION VALUE=\"$db_name\">".$db_name;
}
?>

<html>
   
<body background="lgrey013.jpg" bgcolor="#FAEBD7">
        <label> Databases Available </label>
      <SELECT NAME=db_name>

          <OPTION VALUE=0>Choose
<?=$options?>
</SELECT>

    </body>

</html>

iam only up to these .i dont have any idea regarding how to choose the selected database name in mysql and list the tables already existing.Plz help me

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