De PHPhulp kroeg
Probleempje met color switch!
Dit is wat ik normaal gebruik:
Code (php)
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$switch_style[0] = '#ffffff';
$switch_style[1] = '#edf2f9';
$style = 0;
While ($row = mysql_fetch_assoc($query))
{
$style = ($style == 0) ? 1 : 0;
echo '<div style=background-color: "'.$switch_style[$style].'"</div>';
$switch_style[1] = '#edf2f9';
$style = 0;
While ($row = mysql_fetch_assoc($query))
{
$style = ($style == 0) ? 1 : 0;
echo '<div style=background-color: "'.$switch_style[$style].'"</div>';
Maar nu is het overzicht door iemand anders gemaakt en krijg ik het niet voor elkaar doordat de opbouw anders is, het werkt met functions en total overzicht opbouwen:
Code (php)
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<?php
function getName($id) {
$sql = "SELECT naam
FROM
leden
WHERE id = '$id'";
//query uitvoeren
$sql_res = mysql_query($sql) or die(mysql_error());
while(list($naams)=mysql_fetch_array($sql_res))
{
$naam = $naams;
}
return $naam;
}
function buildResult($id) {
?>
<div class="items-row-box" style="background-color: <?php echo $switch_style[$style];?>">
function getName($id) {
$sql = "SELECT naam
FROM
leden
WHERE id = '$id'";
//query uitvoeren
$sql_res = mysql_query($sql) or die(mysql_error());
while(list($naams)=mysql_fetch_array($sql_res))
{
$naam = $naams;
}
return $naam;
}
function buildResult($id) {
?>
<div class="items-row-box" style="background-color: <?php echo $switch_style[$style];?>">
Hoe zou ik het nu kunnen maken?
Gewijzigd op 13/02/2014 11:43:40 door Kees Mulder
Code (php)
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<?php
function switchColor() {
$switch_style[0] = '#ffffff';
$switch_style[1] = '#edf2f9';
$style = 0;
while(list($style)=mysql_fetch_array($sql_res))
{
$style = ($style == 0) ? 1 : 0;
}
return $style;
}
function buildResult($id) {
?>
<div class="items-row-box" style="background-color: <?php echo switchColor(); ?>">
function switchColor() {
$switch_style[0] = '#ffffff';
$switch_style[1] = '#edf2f9';
$style = 0;
while(list($style)=mysql_fetch_array($sql_res))
{
$style = ($style == 0) ? 1 : 0;
}
return $style;
}
function buildResult($id) {
?>
<div class="items-row-box" style="background-color: <?php echo switchColor(); ?>">
Leesvoer: http://www.php.net/manual/en/language.variables.scope.php
Maar dan wel op de juiste plaats.
Gewijzigd op 13/02/2014 13:00:31 door - SanThe -
Toevoeging op 13/02/2014 14:57:31:
Ik heb net dit nog geprobeerd maar krijg dan overal 'array' te staan:
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<?php
function getColor()
{
$sql = "SELECT COUNT(*)
FROM leden
";
$switch_style[0] = '#ffffff';
$switch_style[1] = '#edf2f9';
$style = 0;
$sql_res = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_assoc($sql_res))
{
$style = ($style == 0) ? 1 : 0;
$switch_style[$style];
return $switch_style;
}
}
function buildResult($id) {
?>
<div class="items-row-box" style="background-color: <?php echo getColor(); ?>">
function getColor()
{
$sql = "SELECT COUNT(*)
FROM leden
";
$switch_style[0] = '#ffffff';
$switch_style[1] = '#edf2f9';
$style = 0;
$sql_res = mysql_query($sql) or die(mysql_error());
while($row = mysql_fetch_assoc($sql_res))
{
$style = ($style == 0) ? 1 : 0;
$switch_style[$style];
return $switch_style;
}
}
function buildResult($id) {
?>
<div class="items-row-box" style="background-color: <?php echo getColor(); ?>">
Gewijzigd op 13/02/2014 15:53:34 door Kees Mulder
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$switch_style[0] = '#ffffff';
$switch_style[1] = '#edf2f9';
$style = 0;
While ($row = mysql_fetch_assoc($query))
{
$style = ($style == 0) ? 1 : 0;
echo '<div style="background-color: '.$switch_style[$style].'"></div>';
}
$switch_style[1] = '#edf2f9';
$style = 0;
While ($row = mysql_fetch_assoc($query))
{
$style = ($style == 0) ? 1 : 0;
echo '<div style="background-color: '.$switch_style[$style].'"></div>';
}
Gewijzigd op 13/02/2014 15:55:54 door - SanThe -
Code (php)
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$switch_style[0] = '#ffffff';
$switch_style[1] = '#edf2f9';
$style = 0;
While ($row = mysql_fetch_assoc($query))
{
$style = ($style == 0) ? 1 : 0;
echo '<div style="background-color: '.$switch_style[$style].'">Hier wat je er in wilt hebben</div>';
}
$switch_style[1] = '#edf2f9';
$style = 0;
While ($row = mysql_fetch_assoc($query))
{
$style = ($style == 0) ? 1 : 0;
echo '<div style="background-color: '.$switch_style[$style].'">Hier wat je er in wilt hebben</div>';
}