Question

我想在我的phpmyadmin表中获取特定值的ID。因此，我得到了一个以'id_cal'作为AI的表。 id，“ mois”代表带有数字的月份（例如1月为1），“ annee”代表年份。（请参阅calendar table）

我正在尝试为月份和年份设置php变量，如果它们与当前月份和年份匹配，我想获取此特定ID。

我在遇到麻烦的地方注释了php代码，这里是：

<?php

include_once('config.php');

$m = idate('n');
$y = idate('Y');

echo $m; echo "\t"; echo $y; echo "<br>"; echo "<br>";  // The result of this is 7 2019

$reponse = $bdd->query('SELECT * FROM calendrier');

while($donnees= $reponse->fetch()){ 
    $mois = $donnees['mois'];
    $year = $donnees['annee'];
    $id_cal = $donnees['id_cal'];
    echo $id_cal;
    echo "\t";
    echo $mois;
    echo "\t";
    echo $year;
    echo "<br>";
}

// What I am trying to do :

if (($m = $mois) && ($y = $year)){  // If the month and the year are the current month/year
    $i = $id_cal;                   // I want to put the id refering to the current month/year (in my phpmyadmin table) into a new variable
    echo "<br>";                    // and echo this variable (or use it in other ways)
    echo $i;                        // BUT what I am echoing is 24 representing the number of values in my array
}                                   // How can I only get 7 ? (in this exemple, since we are the 7/2019)

这是我在本地主机中得到的内容：echo

我真的不明白为什么我没有7。

此外，我尝试了此方法，而不是我的一段时间：

$donnees= $reponse->fetch();    
$mois = $donnees['mois'];
$year = $donnees['annee'];
$id_cal = $donnees['id_cal'];

// But in this cas I am having $i = 1, so it's the same problem.

在此先非常感谢您的回复，对此我感到很努力。

Answer 1

这是因为在您的while语句的每次迭代中，id_cal被新值id_cal覆盖。

要获得所需的结果，可以将if放在while语句中...

while($donnees= $reponse->fetch()){ 
    $mois = $donnees['mois'];
    $year = $donnees['annee'];
    $id_cal = $donnees['id_cal'];
    echo $id_cal;
    echo "\t";
    echo $mois;
    echo "\t";
    echo $year;
    echo "<br>";
    if (($m == $mois) && ($y == $year)){  
        $my_var_to_use_elsewhere = $id_cal;                 
    }   
}

echo "<br>";                   
echo $my_var_to_use_elsewhere;

PHP：如何获取数组的特定值

1 个答案: