的PHP
我的案子有问题,陈述。我正在尝试搜索2年之间的图书,但遇到麻烦,我可以完美地使用此代码搜索一年,但尝试两年都无法正常工作。我确实知道我很可能会以错误的方式获得期望的结果,但是任何帮助将不胜感激。
我也收到错误通知:未定义的变量:Year1为最后一种情况的else部分。谢谢。
如果Year和Year1具有值,则在Year仅具有值的那两年就应该查找该年份的书。
<?php
include 'header.php';
include 'searchscript.php';
$sql = "SELECT DISTINCT bk.title AS Title, bk.bookid AS BookID, bk.year AS Year, bk.publisher AS Publisher, aut.authorname AS Author
FROM book bk
JOIN book_category bk_cat
ON bk_cat.book_id = bk.bookid
JOIN categories cat
ON cat.id = bk_cat.category_id
JOIN books_authors bk_aut
ON bk_aut.book_id = bk.bookid
JOIN authors aut
ON aut.id = bk_aut.author_id";
if(isset($_GET['searchInput'])){
$input = $_GET['searchInput'];
$input = preg_replace('/[^A-Za-z0-9]/', '', $input);
}
if (isset($input)){
$getters = array();
$queries = array();
foreach ($_GET as $key => $value) {
$temp = is_array($value) ? $value : trim($value);
if (!empty($temp)){
if (!in_array($key, $getters)){
$getters[$key] = $value;
}
}
}
if (!empty($getters)) {
foreach($getters as $key => $value){
${$key} = $value;
switch ($key) {
case 'searchInput':
array_push($queries,"(bk.title LIKE '%$searchInput%'
|| bk.description LIKE '%$searchInput%' || bk.isbn LIKE '%$searchInput%'
|| bk.keywords LIKE '%$searchInput%' || aut.authorname LIKE '%$searchInput%')");
break;
case 'srch_publisher':
array_push($queries, "(bk.publisher = '$srch_publisher')");
break;
case 'srch_author':
array_push($queries, "(bk_aut.author_id = '$srch_author')");
break;
case 'srch_category':
array_push($queries, "(bk_cat.category_id = '$srch_category')");
break;
**case 'Year' && 'Year1':
if("$Year1" ==""){
array_push($queries, "(bk.year = '$Year')");
} else {
array_push($queries, "(bk.year BETWEEN '$Year' AND '$Year1')");
}
break;**
}
}
}
if(!empty($queries)){
$sql .= " WHERE ";
$i = 1;
foreach ($queries as $query) {
if($i < count($queries)){
$sql .= $query." AND ";
} else {
$sql .= $query;
}
$i++;
}
}
$sql .= " GROUP BY bk.title ORDER BY bk.title ASC";
}else{
$sql .= " GROUP BY bk.title ORDER BY bk.title ASC";
}
$rs = mysql_query($sql) or die(mysql_error());
$rows = mysql_fetch_assoc($rs);
$tot_rows = mysql_num_rows($rs);
?>
最佳答案
您的代码:
foreach($getters as $key => $value)
switch ($key) {
case 'Year' && 'Year1':
if("$Year1" ==""){
array_push($queries, "(bk.year = '$Year')");
} else {
array_push($queries, "(bk.year BETWEEN '$Year' AND '$Year1')");
}
break;
}
}
显示两个问题:
case
语句不能这样工作。您不能像使用if()
语句那样使用布尔运算符。 (see manual)您不能指望
$key
中的迭代器变量foreach($getters as $key=>$value)
同时包含两个值,这是通过说'Year' && 'Year1'
来暗示的!要解决这些问题,您可以执行以下操作:
foreach($getters as $key => $value)
switch ($key) {
case 'Year':
if($getters["Year1"] ==""){
array_push($queries, "(bk.year = '{$value}')");
} else {
array_push($queries, "(bk.year BETWEEN '{$value}' AND '{$getters['Year1']}')");
}
break;
}
}
在这种情况下,当
foreach($getters)
按下键'Year'
时,将执行该块。现在,if
语句通过直接访问数组中的值而不是查看迭代器变量来正确处理'Year1'
。