SQL LIKE查询失败-预备语句中出现致命错误

对于

LIKE

SELECt ARTICLE_NO FROM AUCTIONS1 WHERe upper(ARTICLE_NAME) LIKE CONCAt('%', ?, '%')

对于表名，将表名作为参数是一种极其糟糕的做法。

如果出于某些原因仍然需要执行此操作，则需要在准备查询之前将其嵌入查询文本中：

$countQuery = "SELECt ARTICLE_NO FROM $table_name WHERe upper(ARTICLE_NAME) LIKE CONCAt('%', ? ,'%')";if ($numRecords = $con->prepare($countQuery)) {    $numRecords->bind_param("s", $brand);    $numRecords->execute();    $data = $con->query($countQuery) or die(print_r($con->error));    $rowcount = mysql_num_rows($data);    $rows = getRowsByArticleSearch($query, $table, $max);    $last = ceil($rowcount/$page_rows);}