Question

很抱歉这个令人困惑的标题。我很困惑。

所以我要做的就是从我拥有的MySQL数据库打印结果。我的数据库中有一个复选框值为“是”，我希望在打印结果时将其替换为其他单词。

我尝试了不同的方法，但所有这些都打破了页面，因为我是新手，并且不知道我在做什么。

到目前为止，这是我的代码（仅提供我认为相关的代码）：

$keyword= "";
if (isset($_POST["keyword"])) {
    $keyword = ($_POST["keyword"]);
}

$results = mysqli_query($con, "SELECT * FROM pcdata WHERE name LIKE '$keyword' LIMIT 0, 25");

if (!$results) {
    echo "Not found...";
} else {
    echo "Found...<br>";
}

while ($row = mysqli_fetch_array($results)) {
    echo "<br>";
    echo "Name: " . $row['name'] . "<br>";
    echo "Model: " . $row['model'] . "<br>";
    echo "Operating system: " . $row['model'] . "<br>";
    echo "Type of computer: " . $row['pctype'] . "<br>";
    echo "Other information: " . $row['info'] . "<br>";
    echo "Need help ASAP: " . $row['help'] . "<br>";                    
}

Answer 1

为什么不在你的内心尝试一个简单的 if ：

$myvariable='';
    if($row['help']='yes'){
        $myvariable='put_something_here';
    }

在你的回声中就是这样做：

 echo "Need help ASAP: " . $myvariable . "<br>";

或三元解决方案：

$row['help'] == 'yes' ? 'put_something_here' : 'what_do_you_want_to_print_if_it_is_not_yes'

Answer 2

试试这段代码：

$keyword= "";
    if (isset($_POST["keyword"]))
    $keyword=($_POST["keyword"]);
$results=mysqli_query($con,"
SELECT * 
FROM pcdata 
WHERE name LIKE '$keyword' LIMIT 0,25");

    if (!$results)  {
        echo "Not found...";
            }
    else    {
        echo "Found...<br>";
        }

    while ($row = mysqli_fetch_array($results))
        {
            echo "<br>";
            echo "Name: " . $row['name'] . "<br>";
            echo "Model: " . $row['model'] . "<br>";
            echo "Operating system: " . $row['model'] . "<br>";
            echo "Type of computer: " . $row['pctype'] . "<br>";
            echo "Other information: " . $row['info'] . "<br>";
            echo "Need help ASAP: ";
            if ($row['help'] === 'yes'){
                echo 'YES';
            } else {
                echo 'NO';
            }
            echo '<br>';
        }

我们会检查$row['help']的值，如果是，那么＆＃34;是＆＃34;打印＆＃39;是＆＃39;，如果是其他 - 打印＆＃39;否＆＃39;

Answer 3

你也可以使用select语句结合Case语句，这将根据需要尝试以下代码

$keyword= "";

if (isset($_POST["keyword"])) 
{
  $keyword = ($_POST["keyword"]);

}
//used different variable to build query

$selectquery="SELECT id,name,model,pctype,info CASE WHEN help='yes' THEN 'Your Yes String' WHEN help='no' THEN 'Your No String' else 'nothing' END as help FROM pcdata where name like '$keyword'" ;
//passed $selectquery to mysqli_qery
$results = mysqli_query($con, $selectquery);

if (!$results) {
 echo "Not found...";
} else {
echo "Found...<br>";
}

while ($row = mysqli_fetch_array($results)) {
echo "<br>";
echo "Name: " . $row['name'] . "<br>";
echo "Model: " . $row['model'] . "<br>";
echo "Operating system: " . $row['model'] . "<br>";
echo "Type of computer: " . $row['pctype'] . "<br>";
echo "Other information: " . $row['info'] . "<br>";
echo "Need help ASAP: " . $row['help'] . "<br>";                    
}

打印出搜索结果时如何替换mysqli_fetch_array上的'value'？

3 个答案: