我想选择最佳查找结果。我收到一个错误: 在这里,我在下面的文件中看到HTML表单操作和数据库连接,请检查一下。错误消息也提到了这一部分。我正在使用php 7.2而不是一些问题
警告:count():参数必须是实现的数组或对象
在第64行的D:\ xammp \ htdocs \ search \ index.php中是可计数的
警告:在第37行的D:\ xammp \ htdocs \ search \ index.php中使用未定义的常量city_id-假定为“ city_id”(这将在以后的PHP版本中引发错误) >
帮我...错误消息。
PHP:
<?php
$i = 1;
if (count($searchdata) > 0) {
foreach ($searchdata as $places) {
echo '<tr>';
echo '<th>' . $i . '</th>';
echo '<td>' . $places['city'] . '</td>';
echo '<td>' . $places['visiting_place'] . '</td>';
echo '<td>' . $places['history'] . '</td>';
echo '</tr>';
$i++;
}
} else {
echo '<td colspan="4">No Search Result Found.</td>';
}
?>
我正在使用PHP,例如:
<?php
$searchdata = [];
$keyword = '';
if (isset($_POST['search'])) {
$city = $_POST['city'];
$keyword = $_POST['keyword'];
$searchdata = $model->getVisitinPlaceData($city, $keyword);
}
?>
数据库连接和表格数据获取
<?php
class Db {
private $hostname = 'localhost';
private $username = 'root';
private $password = '';
private $database = 'test';
private $conn;
public function __construct() {
$this->conn = mysqli_connect($this->hostname, $this->username, $this->password, $this->database);
if(!$this->conn) {
echo 'Database not connected';
}
}
public function getTouristCity(){
$query = "SELECT * FROM tourist_city WHERE is_enabled = '1'";
$result = mysqli_query($this->conn, $query);
return $result;
}
public function getVisitingPlaces(){
$query = "SELECT * FROM visiting_places WHERE is_enabled = '1'";
$result = mysqli_query($this->conn, $query);
return $result;
}
public function getVisitinPlaceData($cityid, $keyword){
$sWhere = '';
$where = array();
if($cityid > 0) {
$where[] = 'V.city_id = '.$cityid.' AND V.is_enabled = "1"';
}
if($keyword != '') {
$keyword = trim($keyword);
$where[] = "( V.visiting_place LIKE '%$keyword%' OR V.history LIKE '%$keyword%' OR C.city LIKE '%$keyword%' )";
}
$sWhere = implode(' AND ', $where);
if($sWhere) {
$sWhere = 'WHERE '.$sWhere;
}
if(($cityid > 0) || ($keyword != '')) {
$query = "SELECT * FROM visiting_places AS V JOIN tourist_city AS C ON C.city_id = V.city_id $sWhere ";
$result = mysqli_query($this->conn, $query);
return $result;
}
}
}
?>
html表单操作
<form action="" method="post" >
<div class="col-sm-3">
<select name="city" class="form-control">
<option value="0">Select City</option>
<?php foreach($turistCity as $city) {
$checked = ($_POST['city'] == $city[city_id])? 'selected' : '';
echo '<option value="'.$city[city_id].'" '.$checked.'>'.$city[city].'</option>';
}
?>
</select>
</div>
<div class="col-sm-3">
<input type="text" name="keyword" placeholder="Keword" value="<?php echo $keyword;?>" class="form-control">
</div>
<button name="search" type="submit" class="btn btn-primary">Search</button>
</form>
答案 0 :(得分:0)
只需将其包装在检查它是否可计数的条件中即可。
$("body").on('click', '#kodeSparepart', function (event) {
var price = $(this).children('option:selected').data('price');
$('#hargaJualTransaksi').val(price);
}
注意:除非明确定义,否则您永远都不要假设任何事情都是正确的。 // PHP >= 7.3
if(is_countable($searchdata)) {
// Do something
}
// PHP >= 7.1
if(is_iterable($searchdata)) {
// Do something
}
表示期望它确实是一个数组或可以计数的东西。如果情况并非总是如此,那么在继续之前,您需要进行很多检查。
编辑:在这种情况下,对于PHP> = 7.1使用count($something)
或对于> = 7.3使用is_iterable()
。上面的代码段已更新
答案 1 :(得分:0)
有时您无法在使用count的地方包装所有子句,或者使用多个count的情况,只需要创建不可数的空数组,只需在实际代码之前添加即可。
if(!is_countable($your_array))$your_array = Array();
答案 2 :(得分:-1)
PHP 7.3给出了以下错误: //如果请求的页面不存在。 if($ elements ['page']> count($ elements ['pages'])){
将其更改为: if(is_countable($ elements ['page'])){
解决了!