Question

我正在使这个人搜索功能构建jquery（json）和php，从搜索栏发送到php文档，但它的行为不像我想要的那样。

我不知道mysql服务器或php代码是否有问题，但是当我执行我的搜索脚本时，它不会返回（示例）上的值：“e”，“s”或“r”但是如果我输入一个人的全名，或只是一个“a”，它返回所有带有“a”的名字。

这里有一小段我的剧本：

    $string = trim(mysql_real_escape_string($_POST['string']));

    $sql = "SELECT * FROM users WHERE firstName LIKE '%$string%' OR lastName LIKE '%$string%' ORDER BY firstName ASC";
    foreach($db->query($sql) as $row) {
    //usage of $row like: $name = $row['firstName']." ".$row['lastName'];
    }

更新
继承整个剧本：

    $sql = "SELECT * FROM users WHERE firstName like '%$string%' OR lastName like '%$string%' ORDER BY firstName ASC";
    foreach($db->query($sql) as $row) {

    $name = $row['firstName']. " ".$row['lastName'];//combing firstname with lastname
    //replacing æ,ø,å,Æ,Ø,Å with html friendly charcodes
    $remove = array("æ", "ø", "å", "Æ", "Ø", "Å"); 
    $with = array("&aelig;", "&oslash;", "&aring;", "&AElig;", "&Oslash;", "&Aring;");
    $fname = str_replace($remove, $with, $name);

    //composing a answer to the user.
    $stringUser[] = "<li class='searchResultHit'><div class='searchResultLeft'><img src='".$row['picture']."'></div><div class='searchResultCenter'><span class='searchResultContent'><a href='user.php?id=".$row['id']."'>".$fname."</a></span></div></li>";
    }

php + mysql - 喜欢不显示所有值

0 个答案: