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Php函数问题_Php - Fatal编程技术网

Php函数问题

Php函数问题,php,Php,我对函数有问题。我在config/funkcije.php中编写它们。在文件夹config/I中有db.php连接到数据库等。。。当我在浏览器config/funkcije.php中打开时,该页面上没有显示任何内容,但它应该会回显数据库中的一些结果 这是我的职责: include 'db.php'; function prikaz_posebne_ponude() { $sql = "SELECT * FROM posebna_ponuda ORDER BY id DESC LIMI

我对函数有问题。我在config/funkcije.php中编写它们。在文件夹config/I中有db.php连接到数据库等。。。当我在浏览器config/funkcije.php中打开时,该页面上没有显示任何内容,但它应该会回显数据库中的一些结果

这是我的职责:

include 'db.php';
function prikaz_posebne_ponude()
{

    $sql  = "SELECT * FROM posebna_ponuda ORDER BY id DESC LIMIT 1";
    $sql1 = $mysqli->query("$sql");
    while ($row = $sql1->fetch_assoc()) {
        $glavni_naziv = $row[$lang];
        if (empty($glavni_naziv)) {
            echo '';
        } else {

            echo "<div class='row-banner'>";
            echo "<h3><span>" . $langArray['rezervacija_smjestaja'] . "</span></h3>";
            echo "<p>" . $glavni_naziv . "</p>";
            echo "</div>";

        }
    }
}
global $mysqli;
global $langArray;
global $lang;
但是当我删除最后一行上的函数prikaz_posebne_ponude{and}时,一切都很好。为什么会发生这种情况?

您定义了一个函数,但从未调用过它。函数是可重用的代码片段,但要执行包含的语句,必须如下调用函数:

prikaz_posebne_ponude();
您还需要告诉PHP函数中的某些变量是全局变量:

include 'db.php';
function prikaz_posebne_ponude()
{

    $sql  = "SELECT * FROM posebna_ponuda ORDER BY id DESC LIMIT 1";
    $sql1 = $mysqli->query("$sql");
    while ($row = $sql1->fetch_assoc()) {
        $glavni_naziv = $row[$lang];
        if (empty($glavni_naziv)) {
            echo '';
        } else {

            echo "<div class='row-banner'>";
            echo "<h3><span>" . $langArray['rezervacija_smjestaja'] . "</span></h3>";
            echo "<p>" . $glavni_naziv . "</p>";
            echo "</div>";

        }
    }
}
global $mysqli;
global $langArray;
global $lang;
您定义了一个函数,但从未调用它。函数是可重用的代码片段,但要执行包含的语句,必须如下调用函数:

prikaz_posebne_ponude();
您还需要告诉PHP函数中的某些变量是全局变量:

include 'db.php';
function prikaz_posebne_ponude()
{

    $sql  = "SELECT * FROM posebna_ponuda ORDER BY id DESC LIMIT 1";
    $sql1 = $mysqli->query("$sql");
    while ($row = $sql1->fetch_assoc()) {
        $glavni_naziv = $row[$lang];
        if (empty($glavni_naziv)) {
            echo '';
        } else {

            echo "<div class='row-banner'>";
            echo "<h3><span>" . $langArray['rezervacija_smjestaja'] . "</span></h3>";
            echo "<p>" . $glavni_naziv . "</p>";
            echo "</div>";

        }
    }
}
global $mysqli;
global $langArray;
global $lang;

在本节末尾添加:;你会看到结果的。
函数是一段代码,需要时可以随时调用。

在这个prikaz_posebne_ponude的末尾添加;你会看到结果的。
函数作为一段代码,在您需要时随时调用。

首先,在函数外部定义的函数中使用的变量需要是全局变量:

function prikaz_posebne_ponude()
{

    global $mysqli;
    global $langArray;
   ....
}
然后,您需要调用函数以使其运行:

prikaz_posebne_ponude();

首先,在函数外部定义的函数中使用的变量需要是全局变量:

function prikaz_posebne_ponude()
{

    global $mysqli;
    global $langArray;
   ....
}
然后,您需要调用函数以使其运行:

prikaz_posebne_ponude();

您可以将必要的变量解析到函数中:

function function_name ($con,$LangArray,$Lang){
  /*
     Manipulate here 
  */
}
或者让它们全球化:

function function_name(){
  global $mysqli;
  global $langArray;
  global $Lang;
}
然后打电话:

 $LangArray = array("test","test_1");
 $Lang = "Value";
 $mysqli = new mysqli("127.0.0.1","user","pass","db");
 function_name($mysqli,$LangArray,$Lang);

您可以将必要的变量解析到函数中:

function function_name ($con,$LangArray,$Lang){
  /*
     Manipulate here 
  */
}
或者让它们全球化:

function function_name(){
  global $mysqli;
  global $langArray;
  global $Lang;
}
然后打电话:

 $LangArray = array("test","test_1");
 $Lang = "Value";
 $mysqli = new mysqli("127.0.0.1","user","pass","db");
 function_name($mysqli,$LangArray,$Lang);

专业提示:用英语命名变量/函数您需要将$mysqli变量设置为全局:全局$mysqli;,从外观上看也是$langArray。函数很可能没有被调用。$lang中有什么,从哪里获得$langArray?专业提示:用英语命名变量/函数您需要将$mysqli变量设置为全局:全局$mysqli;,从外观上看,$langArray也是如此。该函数很可能没有被调用。在$lang中有什么,您从哪里获得$langArray?我调用了它,但仍然没有发生任何事情!:S@user3002173别忘了添加全局$mysqli;全球美元阵列;全球$朗;我调用了它,但还是什么都没发生!:S@user3002173别忘了添加全局$mysqli;全球美元阵列;全球$朗;感谢你,伙计,这解决了我的问题谢谢,伙计,这解决了我的问题