Php Mysql SELECT 查询 1 列等于 1 个变量

Question

我已经在国内外抛出了 4 个多小时的帖子，似乎错过了一件简单的事情。

我正在尝试让几个用户将他们的“新闻”上传到 MYSQL。
但我只想显示带有附加到该行的登录用户名(userpost)的“新闻”。
$current 是登录者的用户名，有效。
示例 A 不会过滤掉不包含 $current 用户的行。
示例 B 未提供任何输出

所以我尝试了两种方法:

$result = mysqli_query($con,"SELECT * FROM images_tbl");
//echo $current . "2" . $current;
while($row = mysqli_fetch_array($result)) {
    if ($row['userpost'] = '.$current.') {
    $num = 0;
    $num = $num + 1;
    $pic.$num = $row['images_path'];
    $h1 = $row['hlone'];

和B:

 $result = mysqli_query($con,"SELECT * FROM images_tbl WHERE (userpost = '.$current.')");
    echo $current . "2" . $current;
    while($row = mysqli_fetch_array($result)) {
      echo $row['hlone'] . " " . $row['images_path'];
      echo "<img src=\"" .$row['images_path']. "\">";
    }

27、images/08-10-2014-1412752801.jpg(images_path)、2014-10-08、Headline(hlone)、Headline2、story、testb(userpost)

任何帮助将不胜感激。

Answer 1

将 where 子句添加到查询中

//in situation A
$result = mysqli_query($con,"SELECT * FROM images_tbl where username='".$current."'");
//username is column name for user you might have to change this
while($row = mysqli_fetch_array($result)) {
    echo $row['images_path'];
    echo $row['hlone'];
}

在情况 B 中尝试这个

$result = mysqli_query($con,"SELECT * FROM images_tbl WHERE userpost = '".$current."')");
    echo $current . "2" . $current;
    while($row = mysqli_fetch_array($result)) {
      echo $row['hlone'] . " " . $row['images_path'];
      echo "<img src=\"" .$row['images_path']. "\">";
    }