Question

我有2个数组，我想用一个按钮将它们添加到我的sql表中。我不知道使用带有sql的数组。我读了很多页但却找不到简单的方法。

$name=array("Jack", "John", "Fiona");

$country=array("London", "Greece", "Japan"); 

$entry =' went to school';

Sql表：

**id** **|** **name**  **|**    **entry**            **|**      **country**

**1** Jack **|** Jack went to school   **|** London

**2** John  **|**  John went to school    **|** Greece

**3** Fiona  **|**  Fiona went to school  **|** Japan

我试过这样的事情，但它没有用。此致

$sql = mysql_query("INSERT INTO bilgi (name,entry,country) VALUES ('$name[]','$name[]&$entry','$country[]'");

Answer 1

$name=array("Jack", "John", "Fiona");

$country=array("London", "Greece", "Japan"); 

$entry =' went to school';

$SizeOfName=sizeof($name);

for($i=0;$i<$SizeOfName;$i++)
{
    $Name=$name[$i];
    $Country=$country[$i];
    $Entry=$Name$entry;

    $sql = mysql_query("INSERT INTO bilgi (name,entry,country) VALUES ('$Name','$Entry','$Country')";
}

Answer 2

您应该真正升级到PDO或mysqli。 mysql API现已弃用一段时间，将在下一版本的PHP（7）中删除。

要回答你的实际问题，第一步应该是使数据健全，即：

$data = [
    [
        'name'    => 'Jack',
        'country' => 'London',
    ],
    [
        'name'    => 'John',
        'country' => 'Greece',
    ],
    [
        'name'    => 'Fiona',
        'country' => 'Japan',
    ],
];

这使得处理和维护数据变得更加容易。

接下来，您只需要遍历单个数组：

$entry = ' went to school';

foreach ($data as $item) {
    $sql = mysql_query("INSERT INTO bilgi (name,entry,country) VALUES ('$item[name]','$item[name]$entry','$item[country]'");
}

但是，这仍然使用旧的API，并不会阻止SQL漏洞。你应该做什么（在making a database connection proper之后）：

$entry = 'went to school';

$stmt = $pdo->prepare('INSERT INTO bilgi (name,entry,country) VALUES (:name, :entry, :country');

foreach ($data as $item) {
    $stmt->execute([
        'name' => $item['name'],
        'entry' => $entry,
        'country' => $item['country'],
    ]);
}

我还从输入字段中删除了name，因为这是数据库中重复的错误方法。如果名称发生变化，您会被entry列中的名称所困。

这使用了一个未弃用的API。防止SQL注入漏洞，并且更加理智和可维护。

Answer 3

您可以使用此

for ($i = 0; $i < count($name); $i++) {
   $entryVal = $name[$i].$entry;
   $sql = mysql_query("INSERT INTO bilgi (name,entry,country) VALUES  ('".$name[$i]."','".$entryVal."','".$country[$i]."'");
}

Answer 4

<?php

$name=array("Jack", "John", "Fiona");

$country=array("London", "Greece", "Japan"); 

$entry =' went to school';

for($i=0 ; $i<count($name);$i++){

    $sql = mysql_query("INSERT INTO bilgi (name,entry,country) VALUES ('".$name[$i]."','".$name[$i].$entry."','".$country[$i]."'");
}

Answer 5

您可以在一个SQL中插入：

$name = array("Jack", "John", "Fiona");
$country = array("London", "Greece", "Japan"); 
$entry =' went to school';

$values;
for ($i = 0; $i < count($name); $i++) {
    $values[] = "('$name[$i]', '$name[$i]$entry', '$country[$i]')";
}

$sql = 'INSERT INTO bilgi (name, entry, country) VALUES ';
for ($i = 0; $i < count($values); $i++) {
    $sql .= $values[$i];
    if($i < (count($values) - 1)) $sql .= ", ";
}

$result = mysql_query($sql);

在插入数据库之前，请注意您的阵列。 $ name和$ country的元素数必须相等。

Answer 6

这是您的问题的动态解决方案：

<?php
$name=array("Jack", "John", "Fiona","Mack","Raja","Bohemia");
$country=array("London", "Greece", "Japan"); 
$entry ='went to school';

$bigcount = max(count($name),count($country));

for($i=0;$i<$bigcount;$i++)
{
    $finalValues .= "('$name[$i]','$country[$i]','$entry'),";
}
$final = substr($finalValues,0,-1);
$query = "INSERT INTO bilgi (name,entry,country) VALUES $final";
$sql = mysql_query($query);
?>

Php从数组插入sql

6 个答案: