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digibucc

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    PHP Code help

    in PHP Coding Help

    Posted

    you could do something this for your sql statement, freehand so check it:

     

    <?php
$sets = count($_POST) / 2;
$x = 0;
$sql = "INSERT INTO Persons (FirstName,Age) VALUES";
while ($sets > 0){
$sql .= '('. $_POST['FirstName'. $x]. ','. '('. $_POST['Age'. $x]. '),';
$sets --;
$x++;
}
$sql= substr($sql,0,-1); // trim trailing ","
?>

    PHP Code help

    in PHP Coding Help

    Posted

    post will just send the last box's info with the name. they NEED to have different names in order to be saved as individual values, otherwise you just have multiple input boxes each over-writing the one before it.

     

    that is one reason why it was only saving one entry before, the other being your INSERT statement. both need to reflect multiple names or it won't work.

    PHP Code help

    in PHP Coding Help

    Posted

    yeah that would be why.

     

    you're going to need something like this

    <?php
$sql = "INSERT INTO Persons (FirstName,Age) VALUES($_POST['FirstName1'],$_POST['Age1']),($_POST['FirstName2'],$_POST['Age2']),($_POST['FirstName3'],$_POST['Age3'])";
?>

     

    your statement was only inserting one row, this inserts multiple by means of a VALUES statement:

    INSERT INTO tbl_name (a,b,c) VALUES(1,2,3),(4,5,6),(7,8,9);

    otherwise you would need to do multiple insert statements.

     

    you can build a foreach on your _POST to have it build your sql query, that way it's not trying to insert values that don't exist, and it accounts for all values the user created.

    PHP Code help

    in PHP Coding Help

    Posted

    well all the boxes will have the same name, so they can only have one value

     

     

    instead of:

    <?php
for ($i=1; $i<=$count; $i++) {
echo 'Firstname: <input type="text" name="firstname">';
echo 'Age: <input type="text" name="age">';
}
?>

     

    try:

    <?php
for ($i=1; $i<=$count; $i++) {
echo 'Firstname: <input type="text" name="firstname'. $i. '">';
echo 'Age: <input type="text" name="age'. $i. '">';
}
?>

     

    which will give them each a name that adds a number ($i) at the end, so they all have a different name.

    eg:

    firstname1, age1

    firstname2, age2

    firstname3, age3

     

    however, i don't know about that second form being below the first. if it's saving a single result i guess it works but i've not seen that. i don't see how it can send ANY data as it only has the submit button as it's input...

    Code help

    in PHP Coding Help

    Posted

    for #4, i think it's just that your values need to be enclosed:

      $insertSQL = sprintf("INSERT INTO beheerder (naam, paswoord) VALUES ('%s','%s')",$naam,$paswoord);

    instead of

      $insertSQL = sprintf("INSERT INTO beheerder (naam, paswoord) VALUES (%s, %s)",$naam,$paswoord);

    Code help

    in PHP Coding Help

    Posted

    then yes, the first two i assume are because that if statement is not validating to true. 

     

    i would check the statement, maybe like this:

    <?php
if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1")) { 
//$beheerder = new Beheerder(GetSQLValueString($_POST['TFnaam'], "text"),GetSQLValueString($_POST['confirmpasw'], "text"));
echo 'the statement was true!!';
} else { echo 'the statement was FALSE '; }

$beheerder->setObj($beheerder);
?>

     

     

    i'll update more...

     

    as for the third error, i don't know. i'm not good enough in oop to help with that yet...

     

    for #4, i think it's just that your values need to be enclosed:

      $insertSQL = sprintf("INSERT INTO beheerder (naam, paswoord) VALUES ('%s','%s')",$naam,$paswoord);

    instead of

      $insertSQL = sprintf("INSERT INTO beheerder (naam, paswoord) VALUES (%s, %s)",$naam,$paswoord);

     

    Need help Negating Statement

    in PHP Coding Help

    Posted

    well for example

     

    if ($x == $y)

    will be true if $x is '0' and y is 0, even though one is a string and one an integer it is still the value - 0 - that counts.

     

    === will require the type to be the same as well, to be true.

     

    the !, replaces one of the =

     

    so !=, is the opposite of ==, and !== is the opposite of ===

    query results

    in PHP Coding Help

    Posted

    yeah then you don't want in_array, you want to do the math for it.

     

    here's an example from the man page for arithmetic operators:

     

    <?php
if($n1 % $n2 == 0) echo "n1 is divisible by n2";
?>

    because:

    $a % $b Modulus Remainder of $a divided by $b.

    if the remainder is 0, it divided evenly.

     

    so in your code you could do:

     

    <?php
// set layout vars
$x = 1; // set $x
echo '<table><tr>'; //start the table, and the row.
// Query member data from the database and ready it for display
$sql = mysql_query("SELECT * FROM products");
while($row = mysql_fetch_array($sql)){
$product = $row["product"];
$id =$row["id"];
$price =$row["price"];
if(9 % $x == 0) { // if 9 evenly divides by x
echo '<td>';                   // start a column
}
?>
    <div id="products">
      <form action="" method="POST" name="myform<?php echo $id; ?>" class="myform<?php echo $id; ?>"> 
        <input type="hidden" name="hiddenField" class="hiddenField" value="<?php echo $product; ?>" />
        <input type="hidden" name="hiddenField2" class="hiddenField2" value="<?php echo $id; ?>" />
        <input type="hidden" name="hiddenField1" class="hiddenField1" value="<?php echo $price; ?>" />
      <input type="submit" name="submit" class="submit" value="<?php echo $product; ?>" style="background-color:lightgreen; height:50px; width:100px;">   </form>
    </div>   
<?php
$end = $x + 1:
if(9 % $x !== 0) { // if 9 DOES NOT evenly divide by x
echo '<br>';
} else { // else if it DOES, end the column
echo '</td>';
}
$x ++; // increase x to keep count
} // while loop ends
//now outside of the loop, 
$x = $x - 1; //subtract the last $X++ so you know exactly how many buttons were added
if(9 % $x !== 0) { // if 9 DOES NOT evenly divide by x - end the column as it wouldn't have ended itself
echo '</td>';
}
echo '</tr></table>'; // close the row and table

?>

     

    past that you'd need to ask for math help, as i'm not well enough versed yet. i am interested in how well the above solution works though.

    Need help Negating Statement

    in PHP Coding Help

    Posted

    yes

     

    though that leaves no room for more than 1 result - i assume that's not possible as you are searching for a member by id or some primary key, but just wanted to point it out.

     

    from man page:

    $a !== $b Not identical TRUE if $a is not equal to $b, or they are not of the same type.

    Code help

    in PHP Coding Help

    Posted

    the first two may be your if statement:

     

    if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1")) {

     

    beheerder is set if that's true, and beheerder isn't being set - so maybe that's not evaluating to true as it should?

     

    edit: actually, looking at this i can't tell which file is which. you have the includes and then code below that, is that the file included or just more code?

     

    post each page in it's own code tags, if you would, so we can see what they are :)

    query results

    in PHP Coding Help

    Posted

    do you mean random number of buttons in a column or random order of buttons ?

     

    the start numbers must be the end numbers plus 1, it could be coded that way instead of having two arrays.

    so you would change the end numbers to however many buttons you want in that column, and the start numbers to one after each of those (1 will always be a start number).

     

    for random order of buttons, since you are doing it by row you can't do array_rand, you could add "ORDER BY RAND()" to your select statement. if you don't have many rows that's ok, but it's really not the way to do it. you want to try and not use ORDER BY RAND.

     

    http://www.titov.net/2005/09/21/do-not-use-order-by-rand-or-how-to-get-random-rows-from-table/ this link explains an alternative.

     

    or do i not understand you at all?

    Quick help with small piece of code!

    in PHP Coding Help

    Posted

    yes you can do that, but you'll have to change a lot of double quotes, and they are standard for code like that.

     

    i would do this:

     

    <?php
/*
    Copyright (C) 2009  Murad <Murawd>
					Josh L. <Josho192837>

    This program is free software: you can redistribute it and/or modify
    it under the terms of the GNU General Public License as published by
    the Free Software Foundation, either version 3 of the License, or
    (at your option) any later version.

    This program is distributed in the hope that it will be useful,
    but WITHOUT ANY WARRANTY; without even the implied warranty of
    MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.  See the
    GNU General Public License for more details.

    You should have received a copy of the GNU General Public License
    along with this program.  If not, see <http://www.gnu.org/licenses/>.
*/

// Must rewrite
// single quote, not double after echo
echo ' 
<fieldset>
	<legend>
		Manage User
	</legend>
	<table border='0' style="height:100%;width:570px;border:solid 1px #BBB">

	<tr class="navtitle" style="height:25px">

		<th style="width:auto">Mute A User From Posting</th>

	</tr>

	<tr>

		<td align="center" class="tdbox">

			Search for the profile name you wish to mute.

		</td>

	</tr>

	<tr>

		<td align="center">

			<form method="post" action=''>

				<input type="text" name="name" /> 

				<input type="submit" name="search" value="Search" />

			</form>

		</td>

	</tr>

	<tr style="height:25px">

		<td class="tdbox">

			<span style="float:left">

				<a href="?cype=admin">Back To Administration Control Panel</a>

			</span>

			<span style="float:right">

				<a href="?cype=admin&page=unmuteuser">Unmute User</a>

			</span>

		</td>

	</tr>

</table>

</fieldset>
'; // single quote, not double ending the statement
?>

     

    just change the doubles at the beginning and en of the echo statement to singles. just note again, if you need variables in there you would end the single

     

    so

     

    echo 'the variable ='. $variable. ' and the echo continues...';

     

    you put singles around plain text strings, and close them , with a period after to continue the echo. just review that one line you should get it.

    Quick help with small piece of code!

    in PHP Coding Help

    Posted

    you are echoing with double quotes, and then you use double quotes in your html. that effectively ends the echo and leaves html its trying to parse as php

    see after style, it turns colors? that's because the double quote right there, it thinks you ended the echo statement.

     

    you can just use single quotes but you have to end them to insert variables. that's what i do. or escape your double quotes, with \

     

    so:

     

    <?php
echo "abcd\"e\"fghijk";
?>

     

    or even:

     

    <?php
echo "
<fieldset>
	<legend>
		Manage User
	</legend>
	<table border='0' style=\"height:100%;width:570px;border:solid 1px #BBB\">";

?>

    Get vs Post

    in PHP Coding Help

    Posted

    right but it has to parse that else every time, even though it is not doing ANYTHING.

    what i do, is comment the else:

     

    <?php
if($x == $y) {
do something();
} // else { x did not == y}

?>

     

    you can even comment multiple lines, at least then it's not trying to parse an empty else statement.

    query results

    in PHP Coding Help

    Posted

    correct, you're fetching by row so the foreach won't work right, using your while loop you could do this:

     

    <?php
// set layout vars
$x = 1; // set $x
$start = array ('1', '10', '19'); //columns will be started at each of these numbers
$end = array( '9', '18' ) // set numbers to end columns at
echo '<table><tr>'; //start the table, and the row.
// Query member data from the database and ready it for display
$sql = mysql_query("SELECT * FROM products");
while($row = mysql_fetch_array($sql)){
$product = $row["product"];
$id =$row["id"];
$price =$row["price"];
if (in_array($x, $start) { // if x equals a number in the start array
echo '<td>';                   // start a column
}
?>
    <div id="products">
      <form action="" method="POST" name="myform<?php echo $id; ?>" class="myform<?php echo $id; ?>"> 
        <input type="hidden" name="hiddenField" class="hiddenField" value="<?php echo $product; ?>" />
        <input type="hidden" name="hiddenField2" class="hiddenField2" value="<?php echo $id; ?>" />
        <input type="hidden" name="hiddenField1" class="hiddenField1" value="<?php echo $price; ?>" />
      <input type="submit" name="submit" class="submit" value="<?php echo $product; ?>" style="background-color:lightgreen; height:50px; width:100px;">   </form>
    </div>   
<?php
if (!in_array($x, $end)){ // if $x equals anything OTHER than 9, 18, etc - put in a line break
echo '<br>';
} else { // else if it DOES equal 9, 18 ,etc end the column
echo '</td>';
}
$x ++; // increase x to keep count
} // while loop ends
//now outside of the loop, 
$x = $x - 1; //subtract the last $X++ so you know exactly how many buttons were added
if (!in_array($x, $end)){ // if $x equals anything OTHER than 9, 18, etc - end the column as it wouldn't have ended itself
echo '</td>';
}
echo '</tr></table>'; // close the row and table

?>

     

     

    query results

    in PHP Coding Help

    Posted

    ok...

     

    foreach is a way of handling arrays in php. an array is a group of elements, so a foreach is saying

     

    for EACH element AS $name, do this:

     

    so

    <?php 
$start = array ('1', '10', '19'); //columns will be started at each of these numbers
$end = array( '9', '18' ) // columns will be ended at each of these numbers

$x = 1; // set $x
echo '<table><tr>'; // start the table
foreach ($result as $button) { // i don't know the name of your array of buttons, i called it $result. this takes the array and processes each button according to what we have below.
if (in_array($x, $start) { // if x equals a number in the start array
echo '<td>';                   // start a column
}
echo $button. '<br>'; // output the button
if (in_array( $x, $end) { // if x equals a number in the end array 
echo '</td>'; // end the column
}
$X++; // increase $x by one
} // do it again. 
?>

    the foreach will repeat until it has process each array element.

     

    i forgot an essential part, $x ++; , which tells it to increase x by one. i added it here. i also forgot a bracket. sorry ;(

    query results

    in PHP Coding Help

    Posted

    need a little more info on how you need it formatted. do you just want that div duplicated 9 times? how exactly would you like them to be laid out in columns, ie one cell per button, or is a line break fine?

    it's just formulaic html coding, you just have to find a piece of logic that explains properly what you want to achieve.

     

    for example,

     

    <?php $start = array ('1', '10', '19');
$end = array( '9', '18' )

$x = 1;
echo '<table><tr>';
foreach ($result as $button) {
if (in_array($x, $start) {
echo '<td>';
}
echo $button. '<br>';
if (in_array( $x, $end) {
echo '</td>';
}
?>

    you can do with math, which may be better - but i haven't put much time into php math so i'm not sure how...

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