echo all from table where id=$_post[ID]

[thearts.beach]

$data= "echo * from table1 where id=$_post[ID] "

can anyone see why this line duz not compute please ?

[ryantroop]

:-/

assuming you have connected to the database correctly...

$result = mysql_query("select * from table1 where id=$_POST[ID]");
while ($data = mysql_fetch_array($result))
{
    foreach($data as $k=>$v)
        echo ($k . " = " . $v);
}

or.. something to that effect... but since we are all moving to mysqli, this is already horridly outdated.

Check out http://www.php.net/manual/en/mysqli.quickstart.statements.php

To expand...

When you query with php, you are asking for a "result" object (or, to be syntactically correct, a resource). You then need to iterate through the object/array/resource/whatever in order to actually get your data.

[thearts.beach]

Thanks for the pointers to the new scripts,

Have tryied your soulation on the above and get the error

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/a5153944/public_html/update1phone.php on line 4

so tryied

$result = mysql_query("select * from table1 where id=8");

and got a record set, so my thinking is

this is the inputpage.php form

<FORM method=post action=input1.php>
<P><textarea name=id>input number here</textarea><P>
<INPUT value="Submit Query" type=submit>
</P><P></P><P></P><P>

</FORM>

[thearts.beach]

thanks @ryantroop

but since we are all moving to mysqli, this is already horridly outdated.

Check out http://www.php.net/manual/en/mysqli.quickstart.statements.phpHere

if ($mysqli->connect_errno) {
    echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}



$res = $mysqli->query("SELECT id, area FROM plate1 WHERE id =('$_POST[id]')");
$row = $res->fetch_assoc();


printf("id = %s (%s)\n", $row['id'], gettype($row['id']));
printf("label = %s (%s)\n", $row['area'], gettype($row['area']));