Question

我在这里遇到了一个问题，我在未回答的问题mysql query returning false even when values DO exist in table? Trying to find if not in table?

中进行了描述

基本上，我试图回显一个特殊的图像以及链接IF如果用户名字符串在数组中，这意味着已经开始与它们进行对话。输入搜索栏时会发生这种情况。

我已经成功存储了那些不应该打印或打印特殊图像的用户：

$convoArray = array();

                echo '<form class="usersearchpm" method="post"><input class="searchbarpm" name="searchbarpm"></input></form>';
                $con = mysqli_connect("localhost","username","password","sqlserver");

                    //$num = mysqli_query($con, "SELECT * FROM `pm_messages` WHERE user_from=".$account['id']."");
                $numCon = mysqli_query($con, "SELECT * FROM `conversation` WHERE user_one=".$account['id']."");
                    $numrows = mysqli_num_rows($numCon);
                while ($u = mysqli_fetch_assoc($numCon))
                    {
                    //get other users usernames to echo link
    $getUserTwo = mysqli_query($con, "SELECT * FROM `accounts` WHERE id=".$u['user_two']."");
        $s = mysqli_fetch_assoc($getUserTwo); //s[username] is a user that DOES have convo



                    array_push($convoArray, $s['username']);
                    //echo "<a href='message.php?id={$s['id']}'><li><img class = 'dmCircle' src = '../images/chatCircle.png'/>{$s['username']} </li></a>";
                }

这很有效。搜索出现问题 - 此代码会为每个用户打印convo数组的大小，因为每次for循环都会调用echo -

if(isset($_POST['searchbarpm'])){
//  foreach($convoArray as $name)
//  {
//      echo $name;
//  }
//$sess->getUsers();
    $dbh = mysqli_connect("localhost","username","password","sqlserver");
                    $query = $_POST['searchbarpm'];
                    $q = mysqli_query($dbh, "SELECT * FROM sqlserver.accounts WHERE username LIKE '%".$query."%'");
                    //display all the results
                    while($row = mysqli_fetch_assoc($q)){

//                      $checkConvo = mysqli_query($dbh, "SELECT * FROM sqlserver.conversation WHERE user_one=".$user_id." AND user_two=".$row['id']."");

                        if($row['id']!= $user_id) { //only output users they dont have convo going with because theyre already printed!!!

                        for($i=0;$i<sizeof($convoArray);$i++)
                        {
                            if($row['username']==$convoArray[$i])
                            {
                                echo 'match!!';
                                echo $row['username'];
                                echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/chatCircle.png'/> {$row['username']}</li></a>";
                            }
                            else 
                            {
                                echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/noChatCircle.png'/> {$row['username']}</li></a>";
                            }
                        }

                        }
                    }
}//

此代码确实识别搜索到的用户何时在阵列中并且DOES打印特殊图像，但是它再次执行此操作多次。

我需要帮助，没有这个就无法完成我的项目。在这里挣扎了一个星期。

如何使用特殊图像打印阵列中的用户？或者让搜索完全没有打开它们？

我做错了什么？

Answer 1

使用此[假设$ convoArray包含用户名是其数组值]

if(in_array($row['username'], $convoArray) {
    echo 'match!!';
    echo $row['username'];
    echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/chatCircle.png'/> {$row['username']}</li></a>";
}
else {
    echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/noChatCircle.png'/> {$row['username']}</li></a>";
}

而不是以下代码

for($i=0;$i<sizeof($convoArray);$i++)
{
    if($row['username']==$convoArray[$i])
    {
        echo 'match!!';
        echo $row['username'];
        echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/chatCircle.png'/> {$row['username']}</li></a>";
    }
    else 
    {
        echo "<a href='message.php?id={$row['id']}'><li><img class = 'dmCircle' src = '../images/noChatCircle.png'/> {$row['username']}</li></a>";
    }
}

PHP - 仅当字符串不在数组中时，回显链接ON ON ONCE链接多次回显？

1 个答案: