Question

我正在尝试创建一个基本的“搜索所有匹配的行的表列，并向我显示行”类型搜索。我最终得到的是一个永远重复的查询，结果网页永远不会停止加载。循环的经典示例无法结束。我不知道为什么。我试图让每个小步骤保持自己的功能就像一个好孩子，并重用功能。

一个人应该能够搜索问题或 john 以返回一行 plus 或 minus 看起来像两行，而 test 将返回三行。

请帮忙！

function databaseconnection($sql){
    $usernm="XXXfooXXX";
    $passwd="XXXfooXXX";
    $host="XXXfooXXX";
    $database="contact_info";
    mysql_connect($host,$usernm,$passwd);
    mysql_select_db($database);
    $result = mysql_query ($sql) or die (mysql_error ());
    return $result;
}

function searchtable(){
    echo 'searchtable() <br />';
    if ($_POST['search'] != "" ){
        $search = preg_replace('/[^ \wa-zA-Z0-9_.@()\-+~,?]+/', '', $_POST['search']);
    }
    if ($search){
        $sql = "SELECT * FROM names_numbers WHERE name LIKE \"%{$search}%\"".
            " OR phone_address LIKE \"%{$search}%\"".
            " OR notes LIKE \"%{$search}%\"";
        echo 'attempting sql action<br />';
        echo "$sql <br />";
        while ($row = mysql_fetch_assoc(databaseconnection($sql))){
            print_r($row);
        }
    }
}

function draw_search_form(){
    echo '<form action="./numbers.php" method="post">'."\n";
    echo 'notes:<br>'."\n";
    echo '<input type="text" name="search"><br>'."\n";
    echo '<input type="submit" value="Submit">'."\n";    
    echo '</form>'."\n";   
}

Answer 1

我相信你的while循环，你每次都在执行查询，这会重置你的结果迭代器。当你切换到mysqli或PDO时，你可以重构一次运行查询，然后迭代结果。

$results = databaseconnection($sql);
while($row = mysqli_fetch_assoc($results) {
    ...
}

Answer 2

使用当前的while循环，您将打开一个新的数据库连接，运行查询并反复获取第一行。

通过将查询移出数据库，然后迭代结果

，可以轻松解决此问题

以下是带有更改的代码

function databaseconnection($sql){
    $usernm="XXXfooXXX";
    $passwd="XXXfooXXX";
    $host="XXXfooXXX";
    $database="contact_info";
    mysql_connect($host,$usernm,$passwd);
    mysql_select_db($database);
    $result = mysql_query ($sql) or die (mysql_error ());
    return $result;
}

function searchtable(){
    echo 'searchtable() <br />';
    if ($_POST['search'] != "" ){
        $search = preg_replace('/[^ \wa-zA-Z0-9_.@()\-+~,?]+/', '', $_POST['search']);
    }
    if ($search){
        $sql = "SELECT * FROM names_numbers WHERE name LIKE \"%{$search}%\"".
            " OR phone_address LIKE \"%{$search}%\"".
            " OR notes LIKE \"%{$search}%\"";
        echo 'attempting sql action<br />';
        echo "$sql <br />";
        // Run the query once
        $result = databaseconnection($sql);

        // Now iterate over the results
        while ($row = mysql_fetch_assoc($result)){
            print_r($row);
        }
    }
}

function draw_search_form(){
    echo '<form action="./numbers.php" method="post">'."\n";
    echo 'notes:<br>'."\n";
    echo '<input type="text" name="search"><br>'."\n";
    echo '<input type="submit" value="Submit">'."\n";    
    echo '</form>'."\n";   
}

Answer 3

发布作为答案，以便更容易格式化已更正的功能：

function searchtable(){
    echo 'searchtable() <br />';
    if ($_POST['search'] != "" ){
        $search = preg_replace('/[^ \wa-zA-Z0-9_.@()\-+~,?]+/', '', $_POST['search']);
    }
    if ($search){
        $sql = "SELECT * FROM names_numbers WHERE name LIKE \"%{$search}%\"".
            " OR phone_address LIKE \"%{$search}%\"".
            " OR notes LIKE \"%{$search}%\"";
        echo 'attempting sql action<br />';
        echo "$sql <br />";
        $results = databaseconnection($sql);
        while($row = mysql_fetch_assoc($results)) {
            print_variable($row);
        }
        //while ($row = mysql_fetch_assoc(databaseconnection($sql))){
        //    print_variable($row);
        //}
    }
}

Answer 4

伙计，你正在提供一个阵列，而这将始终保持真实。请改用foreach。

$row = mysql_fetch_assoc(databaseconnection($sql)) foreach ($row as $r){ print_r ($r)//extract associative elements }

php mysql搜索永远持续

4 个答案: