问在PHP中如何将从一个函数的表中获取数据传递给另一个函数？

提问于 2017-11-16 15:39:33

回答 2关注 0查看 428

我有一个名为tab2的MySQL表。它包含姓名和电话字段

创建了一个名字为 gettabledata()：

function gettabledata()

    $data = array();

    $query = mysql_query("SELECT * FROM tab2");

    $row = mysql_fetch_assoc($query);

     //Here I am unable to get that how can I pass the fetched data in row

    //  variable to anywhere this function called.

 //  After update:

     return $row;

if(isset($_POST['action']))

       $gettab = createtable();//calling the function gettabledata

       //Here I want to get the all data from that table

     // After doing Update:

      foreach($gettab as $row){

        echo $row['name'];

        echo $row['phone'];

执行后，得到如下警告：

Warning: Illegal string offset 'name' in E:\xampp\htdocs\practive\csvtest.php on line 29

Notice: Uninitialized string offset: 0 in E:\xampp\htdocs\practive\csvtest.php on line 29

Warning: Illegal string offset 'phone' in E:\xampp\htdocs\practive\csvtest.php on line 30

Notice: Uninitialized string offset: 0 in E:\xampp\htdocs\practive\csvtest.php on line 30

Warning: Illegal string offset 'name' in E:\xampp\htdocs\practive\csvtest.php on line 29

Warning: Illegal string offset 'phone' in E:\xampp\htdocs\practive\csvtest.php on line 30 `

另外在print_r($gettab)，我得到了：

Array ( [name] => [phone] => 0 )

它只是从表中的第一个记录。

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