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 个多小时的大量线程，并且似乎遗漏了一件简单的事情。

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

所以我两个都试过了：

$result = mysqli_query($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'];

和乙：

 $result = mysqli_query($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

在查询中添加 where 子句

//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'];
}

情况B试试这个

$result = mysqli_query($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']. "\">";
    } 

Answer 2

您为什么要尝试使用 PHP 进行过滤。

如果你想过滤当前用户没有写的“新闻”，只需使用 MySQL Where 子句：

// For Example A
$result = mysqli_query($con, "SELECT * FROM images_tbl WHERE userpost != '{$current}'");
while($row = mysqli_fetch_array($result)) {
    $pic = $row['images_path'];
    $h1 = $row['hlone'];
}

// For Example 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']. "\">";
} 

使用 MySQL 的过滤选项很容易。你应该对 MySQL 做更多的研究。