str_replace 函数无法替换多个值

samson 发表于 Dev

参孙

我有一组字符串，需要使用 str)replace 替换它们，因为我们遍历查询，因为查询成功需要替换 PHP 中的每个值。

<?php

include_once("con.php");
$db = new Da();

$con = $db->con();

$lclString = "{{ONE}} {{TWO}} {{THREE}} {{FOUR}} {{FIVE}} {{SIX}}";

 $lclCount = 1;
 if(preg_match_all("/\{\{[^{}]+\}\}/", $lclString, $matches)) {

    foreach ($matches[0] as $match) {

        $lclQuery = "SELECT link, link_value FROM student WHERE link_value = '".$match."'";

        $lclResult = $connString->query($lclQuery);

        while($row = $lclResult->fetch(PDO::FETCH_ASSOC)) {

            $lclNewValue = str_replace($match, $row['link'], $lclString);

            }

        }

        echo json_encode($NewValue);

 } 

?>

这里只发生了第二次更换。

参孙

<?php

include_once("con.php");
$db = new Da();

$con = $db->con();

$lclString = "{{ONE}} {{TWO}} {{THREE}} {{FOUR}} {{FIVE}} {{SIX}}";

 $lclCount = 1;
 if(preg_match_all("/\{\{[^{}]+\}\}/", $lclString, $matches)) {
 $lclNewValue = $lclString;
foreach ($matches[0] as $match) {

    $lclQuery = "SELECT link, link_value FROM student WHERE link_value = '".$match."'";

    $lclResult = $connString->query($lclQuery);

    while($row = $lclResult->fetch(PDO::FETCH_ASSOC)) {

        $lclNewValue = str_replace($match, $row['link'], $lclNewValue);

        }

    }

    echo json_encode($NewValue);

 } 

?>

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