使用count函数确定if语句中显示的内容 [英] Using the count function to determine what displays in if statement
问题描述
我要进行的查询遇到一些问题.我正在做的是尝试为多个页面创建模板，以显示与该页面相关的特定图像.
I am running into some issues with a query I am trying to make. What I am doing is trying to create a template for multiple pages to display specific images that correlate with that page.
我在各个页面上分配变量来定义页面(您可以在代码中使用$page看到该位置).然后在我的数据库中，在solution下，我将特定的记录命名为各个页面名称之一.例如:如果我在数据库列solution下将页面命名为球"，我将命名为几个记录Ball.
I have individual pages where I am assigning a variable to define the page (you can see where I do this in my code with $page). Then within my database, under solution I am naming the specific records one of the individual page names. For example: if I named a page "Ball", under the database column solution, I would name a few records Ball.
然后在查询中，我试图计算存在与$page相匹配的记录.如果记录计数大于0，我想在我的else语句中显示代码.
Then within my query, I am trying to count how many records exist that match $page. If the record count is more than 0, I want to display the code in my else statement.
到目前为止，我的数据库连接正在工作.我没有收到任何打印错误.您可以看到我的echo $solution_count;.这显示为0，但是我的else-statement正在运行，这使0有意义.
As of now, my database connection is working. I am not getting any errors being printed. You can see my echo $solution_count;. This is showing a 0, but my else-statement is running, which makes 0 sense.
我在尝试计算记录数量上有错吗?有人知道为什么这行不通吗?
Am I doing anything wrong with how I am trying to count the records? Does anyone see why this isn't working?
数据库表-显示创建表
projectslider
CREATE TABLE `projectslider` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`solution` varchar(50) NOT NULL,
`image` text NOT NULL,
`alt` text NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=latin1
各个页面上的代码:
<?php
error_reporting(E_ALL);
ini_set('display_errors', 1);
$page = "enclosures";
include_once("projectSlider.php");
?>
母版页-projectSlider.php
error_reporting(E_ALL);
ini_set('display_errors', 1);
$servername = 'localhost';
$username = 'root';
$password = '';
try {
$con = new PDO('mysql:host='.$servername.';dbname=mb', $username, $password);
$con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
//$page = mysql_real_escape_string($page);
//SQL Call
$sql_project = "
SELECT *, COUNT(solution) AS solution_count
FROM projectslider
WHERE solution = '. $page .'
";
if ($project_stmt = $con->prepare($sql_project)) {
$project_stmt->execute();
$project_rows = $project_stmt->fetchAll(PDO::FETCH_ASSOC);
foreach ($project_rows as $project_row) {
$solution_count = $project_row['solution_count'];
echo $solution_count;
$project_solution = $project_row['solution'];
$project_img = $project_row['image'];
$project_alt = $project_row['alt'];
$project_img = '<img class="home-comment-profile-pic" src=" '. $project_img .'" alt="' . $project_alt .'">';
if ($solution_count === 0) {
echo 'No projects found.';
} else {
echo '<section id="solProj">';
echo '<div class="projSlide">';
echo $project_img;
echo '</div>';
echo '</div>';
}
}
}
}
catch(PDOException $e) {
echo "Connection failed: " . $e->getMessage();
}
推荐答案
一个项目有很多滑块，为此，您应该有两个具有关系的表项目和projectslider.
One project have many sliders , for this you should have two tables projects and projectsliders with relationship.
项目表:
CREATE TABLE `projects` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`project_name` varchar(50) NOT NULL
)
projectsliders:
projectsliders:
CREATE TABLE `projectsliders` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`solution` varchar(50) NOT NULL,
`image` text NOT NULL,
`alt` text NOT NULL,
`project_id` int(11),
PRIMARY KEY (`id`)
) ENGINE=InnoDB AUTO_INCREMENT=2 DEFAULT CHARSET=latin1
获取一个项目的ProjectSliders，母版页-projectSlider.php
fetching projectSliders for one project, Master page - projectSlider.php
获取一个项目的projectslider的最佳方法是使用OOP，您可以调用一个方法并传递项目ID，并且method应该返回一个包含该项目的projectsliders的数组，但是我正在改善您的代码.
Best way to fetching projectsliders for one project is to use OOP you can call a method and pass project id and method should return you a array with projectsliders for this project , but i am improving your code.
error_reporting(E_ALL);
ini_set('display_errors', 1);
$servername = 'localhost';
$username = 'root';
$password = '';
try {
$con = new PDO('mysql:host='.$servername.';dbname=mb', $username,
$password);
$con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
//$page = mysql_real_escape_string($page);
//SQL Call
$sql_project = "SELECT * FROM projectsliders ps inner join projects p
on p.id = ps.project_id
WHERE p.project_name = '. $project_page .'";
if ($project_stmt = $con->prepare($sql_project)) {
$project_stmt->execute();
$count = project_stmt->rowCount();
if( $count != 0 ){
$project_rows = $project_stmt->fetchAll(PDO::FETCH_ASSOC);
foreach ($project_rows as $project_row) {
$project_solution = $project_row['solution'];
$project_img = $project_row['image'];
$project_alt = $project_row['alt'];
$project_img = '<img class="home-comment-profile-pic" src=" '.
$project_img .'" alt="' . $project_alt .'">';
echo '<section id="solProj">';
echo '<div class="projSlide">';
echo $project_img;
echo '</div>';
echo '</div>';
}
}else{
echo 'No projects found.';
}
}
}
catch(PDOException $e) {
echo "Connection failed: " . $e->getMessage();
}
各个页面:
error_reporting(E_ALL);
ini_set('display_errors', 1);
$project_page = "enclosures";
include_once("projectSlider.php");
我希望这可以帮助您，享受编码的乐趣.
I hope that this can help you ,enjoying coding.
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