如果选中复选框,如何回显 'checked'
How to echo 'checked' if checkbox is checked
在我的表单的 "edit view" 中,我有一个复选框列表,我想检查 table book_have_languages
中的复选框(注意一本书可以有更多不止一种语言)。
代码:
<?php $query_c = mysqli_query($mysqli, "SELECT id_language, name_language FROM languages ORDER BY name_language;")
or die('error '.mysqli_error($mysqli));
while ($data_c = mysqli_fetch_assoc($query_c)) {
echo "<div class='checkbox'><label><input type='checkbox' name='languages[]' value='$data_c[id_languages]'>$data_c[name_language]</label></div>";
$sql = "SELECT id_book, id_language FROM book_have_languages WHERE id_book=3";
$res = $mysqli->query($sql) or die("error: ".$sql);
$results = array();
while($row = mysqli_fetch_assoc($res)) {
$results[] = $row;
}
foreach ($results as $result){
if($result['id_language'] == $data_c['id_language']){
echo"<div class='checkbox'><label><input type='checkbox' name='languages[]' value='$data_c[id_language]'checked >$data_c[name_language]</label></div>";
}
}
}
?>
结果:
如果我的table是:
id_book | id_language
3 - 2
3 - 5
3 - 6
已检查语言 2、5、6,但它们似乎是重复的。我如何在第一个 echo 中指示只显示 table 中没有的语言?
您正在循环浏览各种语言,而不是浏览一本书的语言。如果您的 table 与描述的一样,$data_idi
应该有多个结果,而您只匹配第一个。您也应该在这些结果中循环。
我是这样解决的:
<?php
$query_c = mysqli_query($mysqli, "SELECT id_language, name_language FROM languages ORDER BY name_language;")
or die('error '.mysqli_error($mysqli));
while ($data_c = mysqli_fetch_assoc($query_c)) {
$sql = "SELECT id_book, id_language FROM book_te_languages WHERE id_book='$_GET[id]'";
$res = $mysqli->query($sql) or die("error: ".$sql);
$results = array();
while($row = mysqli_fetch_assoc($res)) {
$results[] = $row;
}
echo "<div class='checkbox'><label><input type='checkbox' name='languages[]' value='$data_c[id_language]'";
foreach ($results as $result){
if($result['id_language'] == $data_c['id_language']){
echo "checked";
}
}
echo ">$data_c[name_language]</label></div>";
}
?>
在我的表单的 "edit view" 中,我有一个复选框列表,我想检查 table book_have_languages
中的复选框(注意一本书可以有更多不止一种语言)。
代码:
<?php $query_c = mysqli_query($mysqli, "SELECT id_language, name_language FROM languages ORDER BY name_language;")
or die('error '.mysqli_error($mysqli));
while ($data_c = mysqli_fetch_assoc($query_c)) {
echo "<div class='checkbox'><label><input type='checkbox' name='languages[]' value='$data_c[id_languages]'>$data_c[name_language]</label></div>";
$sql = "SELECT id_book, id_language FROM book_have_languages WHERE id_book=3";
$res = $mysqli->query($sql) or die("error: ".$sql);
$results = array();
while($row = mysqli_fetch_assoc($res)) {
$results[] = $row;
}
foreach ($results as $result){
if($result['id_language'] == $data_c['id_language']){
echo"<div class='checkbox'><label><input type='checkbox' name='languages[]' value='$data_c[id_language]'checked >$data_c[name_language]</label></div>";
}
}
}
?>
结果:
如果我的table是:
id_book | id_language
3 - 2
3 - 5
3 - 6
已检查语言 2、5、6,但它们似乎是重复的。我如何在第一个 echo 中指示只显示 table 中没有的语言?
您正在循环浏览各种语言,而不是浏览一本书的语言。如果您的 table 与描述的一样,$data_idi
应该有多个结果,而您只匹配第一个。您也应该在这些结果中循环。
我是这样解决的:
<?php
$query_c = mysqli_query($mysqli, "SELECT id_language, name_language FROM languages ORDER BY name_language;")
or die('error '.mysqli_error($mysqli));
while ($data_c = mysqli_fetch_assoc($query_c)) {
$sql = "SELECT id_book, id_language FROM book_te_languages WHERE id_book='$_GET[id]'";
$res = $mysqli->query($sql) or die("error: ".$sql);
$results = array();
while($row = mysqli_fetch_assoc($res)) {
$results[] = $row;
}
echo "<div class='checkbox'><label><input type='checkbox' name='languages[]' value='$data_c[id_language]'";
foreach ($results as $result){
if($result['id_language'] == $data_c['id_language']){
echo "checked";
}
}
echo ">$data_c[name_language]</label></div>";
}
?>