在PHP上返回整个MYSQL表数据

在PHP上返回整个MYSQL表数据,php,mysql,Php,Mysql,我曾经找到过这个代码 $sql_select = "SELECT * FROM users"; $rs_stuff = select($sql_select); while ($res = mysql_fetch_assoc($rs_stuff)) { echo ($res("Name")."<br />"); } $sql\u select=“select*FROM users”; $rs\u stuff=select($sql\u select); 而($res=mysq

我曾经找到过这个代码

$sql_select = "SELECT * FROM users";
$rs_stuff = select($sql_select);

while ($res = mysql_fetch_assoc($rs_stuff)) {

echo ($res("Name")."<br />");

}
$sql\u select=“select*FROM users”;
$rs\u stuff=select($sql\u select);
而($res=mysql\u fetch\u assoc($rs\u stuff)){
echo($res(“Name”)。“
”); }
它工作得很好,它返回在“Name”列中找到的所有名称,问题是我希望它返回该表中的所有数据,就像我在mysql上键入“SELECT*FROM users”,我对PHP不太了解,我尝试这样做:

echo("<br />\n".$res);
echo(“
\n”。$res);
但是当我试图在页面上运行这个时,我得到了一个空白,上面写着“Array”。。。 是否可以在不将col名称放入php的情况下执行此操作


(对不起,我的英语不是我的主要语言。)

发生这种情况是因为
$res
是一个数组

使用:

或者,如果您想更好地了解它:

echo '<pre>';
  print_r($res);
echo '</pre>';
如果你只有一个结果。我建议您选择第一个选项,在回显字符串之前先看看
$res
的外观;)

无论哪种方式,请找到一个好的PHP教程,因为很明显,您缺乏如何使用和操作PHP代码的基本知识

希望有帮助!:D

$sql\u select=“从用户选择*”;
 <?php
   $servername = "localhost";
   $username = "username";
   $password = "password";
   $dbname = "myDB";

   // Create connection
      $conn = new mysqli($servername, $username, $password, $dbname);
   // Check connection
   if ($conn->connect_error) {
            die("Connection failed: " . $conn->connect_error);
    } 

     $sql = "SELECT * FROM users";
     $result = $conn->query($sql);

    if ($result->num_rows > 0) {
           // output data of each row
    while($row = $result->fetch_assoc()) {
       print_r($row)
    }
   } else {
      echo "0 results";
}
 $conn->close();
?>
$rs\u stuff=select($sql\u select); 回声'//创建表 而($res=mysql\u fetch\u assoc($rs\u stuff)){ 回显'.$res['name'].'; }
如果你在学习
mysqli\uuu
PDO
它们在学习
mysql
时没有意义,因为它们被定价了

试试这个

$sql_select = "SELECT * FROM users";
$rs_stuff = select($sql_select);
$arr = array();
while ($res = mysql_fetch_assoc($rs_stuff)) {
  $arr[] = $res["Name"];
}

echo '<pre>';
print_r($arr);
echo '<pre>';

请将所有行放入一个数组中,然后打印

foreach($res as $key => $value) {
    print $res[$key] . ' => ' . $value;
}
$sql\u select=“select*FROM users”;
$rs\u stuff=select($sql\u select);
$arr=array();
而($res=mysql\u fetch\u assoc($rs\u stuff)){
$arr[]=$res[“Name”];
}
回声';
印刷费($arr);
回声';
试试这个:

function convertArraytoString($array) {
    foreach($array as $key => $value) {
        $return .= $key.': '.$value."<br />\n";
    }
    return $return;
}

希望有帮助:)

它认为这将有助于:


然后,您可以通过echo convertArraytoString($res)转换为数组

先学习基本php
echo
不打印数组。。。使用
print\u r()
var\u dump()
几乎可以,它没有返回列名称,但返回的像“value=>value”我如何修复它以返回“row\u name=>value”?print“Column name is:”$钥匙“关联值为:”$res[$key];
 <?php
   $servername = "localhost";
   $username = "username";
   $password = "password";
   $dbname = "myDB";

   // Create connection
      $conn = new mysqli($servername, $username, $password, $dbname);
   // Check connection
   if ($conn->connect_error) {
            die("Connection failed: " . $conn->connect_error);
    } 

     $sql = "SELECT * FROM users";
     $result = $conn->query($sql);

    if ($result->num_rows > 0) {
           // output data of each row
    while($row = $result->fetch_assoc()) {
       print_r($row)
    }
   } else {
      echo "0 results";
}
 $conn->close();
?>
$sql_select = "SELECT * FROM users";
$rs_stuff = select($sql_select);
$arr = array();
while ($res = mysql_fetch_assoc($rs_stuff)) {
  $arr[] = $res["Name"];
}

echo '<pre>';
print_r($arr);
echo '<pre>';
foreach($res as $key => $value) {
    print $res[$key] . ' => ' . $value;
}
$sql_select = "SELECT * FROM users";
$rs_stuff = mysql_query($sql_select);

while ($res = mysql_fetch_assoc($rs_stuff)) {
    foreach($res as $key => $value) {
        echo $key.': '.$value."<br />\n";
    }
 echo "<br />\n";
}
Name: John<br />
Usernmae: johndoe<br />
Age: 36<br />
<br />
Name: Christian<br />
Username: admin<br />
Age: 46<br />
<br />
function convertArraytoString($array) {
    foreach($array as $key => $value) {
        $return .= $key.': '.$value."<br />\n";
    }
    return $return;
}