【发布时间】:2016-11-21 00:10:33
【问题描述】:
我已经学习 PHP 一个半月了。我尝试创建从find_all_subjects() 获取结果的函数:
function find_all_subjects() {
global $dbconnect;
$q = "SELECT * FROM subjects ORDER BY id ASC";
$subject_set = mysqli_query($dbconnect, $q);
confirm_query($subject_set);
return $subject_set;
}
然后创建一个新函数,将结果循环到表行中。问题是它只循环一个结果......当我使用<ul> 和<li> 执行此操作时,它运行良好,但在使用表格时似乎不起作用。我的表格核心标签在另一个文档中。所以不是这样的……
function navigacija() {
$s = find_all_subjects();
while($subjects = mysqli_fetch_assoc($s)) {
$out = "<tr>";
// Ime Teme
$out .= "<td width='25%' height='40'> <a href=\"admin_content.php?subject=" . urlencode($subjects['id']) . "\">";
$out .= $subjects['menu_name'];
$out .= "</a></td>";
// Vidljiva
if($subjects['visible'] == 1) {
$subjects['visible'] = 'DA';
} else {
$subjects['visible'] = 'NE';
}
$out .= "<td width='25%' height='40'><p align=\"center\">DA / NE";
$out .= "</p></td>";
// Broj Strana
$out .= "<td width='25%' height='40'>";
$pages_set = find_sub_from_pages($subjects['id']);
while($pages = mysqli_fetch_assoc($pages_set)) {
$out .= "<a href=admin_content.php?page=" . $pages['id'] . ">" . $pages['menu_name'] . "</a>";
$out .= "</td>";
}
// Vidljiva
$out .= "<td align=\"center\" width='25%' height='40'>";
$out .= "<img width=\"17px\" height=\"17px\" src=\"st/img/ic-arup.png\">
</img> <img width=\"17px\" height=\"17px\" src=\"st/img/ic-ardown.png\"></img> ";
$out .= "</td>";
$out .= "</tr>";
}
return $out;
}
【问题讨论】:
标签: php loops mysqli while-loop html-table