Retrive The Values From Array

[yangkai9999]

Hello,I can use the following code to get values into array info. I need to present those values at a html form and to make change later. I don't know how to retrive the values from the array into post variables. Can I get example form you?Thank you,code://get the value from table p2 for a specific p_id;$query = mysql_query("SELECT * FROM mytest.p2 WHERE p_id=$chk;");// puts the P2 data into the $info array $info = mysql_fetch_array( $query ); // Print out the contents of the entry Print "<b>P_ID:</b> ".$info['P_ID'] . " <br>"; Print "<b>v1:</b> ".$info['v1'] . " <br>"; Print "<b>V2:</b> ".$info['V2'] . " <br>";

[jeffman]

I don't know how to retrive the values from the array into post variables.Do you really mean $_POST variables? I don't think you do. Maybe you chose the wrong word.Is mytest.p2 really the name of your table? If so, ok. (I've never had a table name with a dot in it, so it looks weird to me.)If p_id is the name of a field, then $info['P_ID'] won't work. You'll need $info['p_id']Query string values need to be in quotation marks. If $chk is a string, then you'll need them. If it is a number type, then you won't (that is, don't write '$chk' as I have it below). Query strings should not end with a semicolon the way they do when you use SQL directly. Try this:mysql_query("SELECT * FROM mytest.p2 WHERE p_id='$chk'");Some implementations require tick marks around the field name. This might work if the above does not:mysql_query("SELECT * FROM mytest.p2 WHERE `p_id`='$chk'");I am concerned that the next two statements use different cases for the field names. It's fine if the field names really are in different cases, but I don't imagine a designer doing that intentionally:Print "<b>v1:</b> ".$info['v1'] . " <br>"; Print "<b>V2:</b> ".$info['V2'] . " <br>";

[jlhaslip]

echo "<table border=\"1\" align=\"center\">";echo "<tr><th>Quantity</th>";echo "<th>Price</th></tr>";while ( $counter <= 100 ) {	echo "<tr><td>";	echo $counter;	echo "</td><td>";	echo $brush_price * $counter;	echo "</td></tr>";	$counter = $counter + 10;}echo "</table>";

that should give you an idea about how to do it.

[jeffman]

Whoa. Are we working on the same project, Jim? I've been confused all week . . .

[yangkai9999]

Whoa. Are we working on the same project, Jim? I've been confused all week . . .

The code I posted is working. I just want to know how to retrive the values in the array (values of P_ID, V1 and V2) and show them at a html form and let user to modify them.Or, may be I should go to another way to do that?Thank you,

[jeffman]

If your query is working, then this part of the code should do what you expect:// Print out the contents of the entry Print "<b>P_ID:</b> ".$info['P_ID'] . " <br>"; Print "<b>v1:</b> ".$info['v1'] . " <br>"; Print "<b>V2:</b> ".$info['V2'] . " <br>";What's happening instead?

[jlhaslip]

I interpreted it as needing a method to print an array of values???

[jeffman]

OH! You want them in a form.echo "<input type='text' name = 'P_ID' value='{$info['P_ID']}'><br>"; echo "<input type='text' name = 'v1' value='{$info['v1']}'><br>"; echo "<input type='text' name = 'V2' value='{$info['V2']}'><br>"; Can you do the rest of the form yourself?

[jeffman]

I interpreted it as needing a method to print an array of values???

So you're showing how a loop makes a table. Got it.

[yangkai9999]

So you're showing how a loop makes a table. Got it.

I got it. thank you.

[yangkai9999]

one more question.I would like to use the code to get a value for radio group. It seems having prob. How to change it?thanks,code:How many?     <input type="radio" <?php if ($info['V4'] == "1") {echo "enabled";} ?> value="1" name="v4">   1  <input type="radio" <?php if ($info['V4'] == "2") {echo "enabled";} ?> value="2" name="v4">   2  <input type="radio" <?php if ($info['V4'] == "3") {echo "enabled";} ?> value="3" name="v4">   3 </p>

[jeffman]

1. There is no enabled property. Enabled is the default. There is a disabled property you could assign the others.2. You can't just put the word "disabled" in the middle of your tag. Something like this will be needed:echo "disabled='disabled'";