select stmt->fetch 在变量 LIKE 语句之后不显示查询

Question

所以我有更多的查询，其中包含取自 $keywords 的值，当我调用 stmt->fetch() 时没有出现这些值我的表有一些结构（列）所以我尝试搜索关键字（列）然后显示titlu(column) 包含特定关键字的查询 ($keywords) 但即使我在关键字列值 top 中有一个示例也没有显示任何内容强>顶部

$stmt = $con->prepare('SELECT keywords FROM stiinta WHERE link = ? LIMIT 1');
$stmt->bind_param('s', $pageid);
$stmt->execute();
$stmt->bind_result($keyword);//variabla pe care o vrei inlocuita prin bind_Result in loc de get_Result
while ($stmt->fetch())    {
   $keyword;    // faci acelasi lucru fara $row 
}
$stmt->close();

测试

$stmt = $con->prepare("SELECT id, titlu, link, poza, alt, keywords, linknews FROM stiinta WHERE   approved='1' AND  replace(replace(replace(titlu, ',', ''), '-', ''), ' ', '') LIKE  ? UNION SELECT id, titlu, link, poza, alt, keywords, linknews FROM travel WHERE   approved='1' AND  replace(replace(replace(titlu, ',', ''), '-', ''), ' ', '') LIKE  ? UNION SELECT id, titlu, link, poza, alt, keywords, linknews FROM stiinta WHERE   approved='1' AND  replace(replace(replace(titlu, ',', ''), '-', ''), ' ', '') LIKE  ? UNION SELECT id, titlu, link, poza, alt, keywords, linknews FROM natura WHERE   approved='1' AND  replace(replace(replace(titlu, ',', ''), '-', ''), ' ', '') LIKE  ? UNION SELECT id, titlu, link, poza, alt, keywords, linknews FROM lifestyle WHERE   approved='1' AND  replace(replace(replace(titlu, ',', ''), '-', ''), ' ', '') LIKE  ? LIMIT 10");
$stmt->bind_param("sssss", $keyword, $keyword, $keyword, $keyword, $keyword);
$stmt->execute();
if(!$stmt->execute()){
    echo "a aparut o eroare";}
$stmt->bind_result($id, $titluKEY, $linkKEY, $pozaKEY, $altKEY, $keywordKEY, $linknewsKEY);//variabla pe care o vrei inlocuita prin bind_Result in loc de get_Result
$stmt->store_result();
if ($stmt->num_rows == 0) {// verificare daca este vrun query
    echo "You did not have any queries to match.<br>";
} else {
$stiintalist = '';
while ($stmt->fetch())    {
    $id;
    $titluKEY;
    $linkKEY;
    $pozaKEY;
    $altKEY; 
    $keywordKEY;
    $linknewsKEY;   // faci acelasi lucru fara $row trb sa fie in concordanta cu ceea ce este in SELECT column pentretu teste a href="/page-stiinta.php?pid='.$linkKEY.'"
    $stiintalist .= '<div id="articol-content-more"><a href="/'.$linknewsKEY.'"><img src="/images/'.$pozaKEY.'.jpg"class="articol-content-more-image" alt="'.$altKEY.'"><p class="articol-content-more-title">'.$titluKEY.'</p></a><span><a class="articol-content-more-afla" href="/'.$linknewsKEY.'">Citește mai multe</a></span><span class="articol-content-more-fl"><div class="fb-share-button" data-layout="button_count" data-href="http://esticurios.ro/'.$linknewsKEY.'"></div></span></div>';        
    }
}
$stmt->free_result();
$stmt->close();

echo $stiintalist;

给我看你没有任何要匹配的查询所以这意味着$stmt->num_rows == 0 是空的，即使我在其他查询中有一些词....

Answer 1

你的列可能是 char 类型，然后尝试添加 wildchar

$stmt = $con->prepare("

    SELECT id, titlu, link, poza, alt, keywords, linknews 
    FROM stiinta 
    WHERE   approved='1' 
      AND  replace(replace(replace(titlu, ',', ''), '-', ''), ' ', '') LIKE  concat('%', ? , '%') 
    UNION 
    SELECT id, titlu, link, poza, alt, keywords, linknews 
    FROM travel 
    WHERE   approved='1' 
      AND  replace(replace(replace(titlu, ',', ''), '-', ''), ' ', '') LIKE  concat('%', ? , '%') 
    UNION 
    SELECT id, titlu, link, poza, alt, keywords, linknews 
    FROM stiinta 
    WHERE   approved='1' 
      AND  replace(replace(replace(titlu, ',', ''), '-', ''), ' ', '') LIKE  concat('%', ? , '%') 
    UNION 
    SELECT id, titlu, link, poza, alt, keywords, linknews 
    FROM natura 
    WHERE   approved='1' 
      AND  replace(replace(replace(titlu, ',', ''), '-', ''), ' ', '') LIKE  concat('%', ? , '%') 
    UNION 
    SELECT id, titlu, link, poza, alt, keywords, linknews 
    FROM lifestyle 
    WHERE   approved='1' 
      AND  replace(replace(replace(titlu, ',', ''), '-', ''), ' ', '') LIKE  concat('%', ? , '%') 
    LIMIT 10
    ");

Answer 2

由于您的 $pageid 可能是一个整数 (?)，您可以尝试将第二行更改为

$stmt->bind_param('i', $pageid);