Question

我使用以下PHP代码来回显数据库中的值：以下内容来自数据库：

BOLERO DI NON AC WHITE BS3
587247
BOLERO DI NON AC SILVER BS3
599524

是否可以像

那样分别存储每一个

var 1=BOLERO DI NON AC WHITE BS3
var 2=587247
var 3=BOLERO DI NON AC SILVER BS3
var 4=599524

PHP代码：

<?php


 $query2="SELECT ABS(price - $new_loan_amount) AS nearest, model, price FROM car_pricing WHERE state = '".$state."' ORDER BY nearest ASC LIMIT 2;
             ";



$result2 = mysqli_query($db,$query2);
 while ($row = mysqli_fetch_assoc($result2)){

      $i=1;
       echo $database_model = $row['model'];
       echo "<br>";
       echo $database_price = $row['price'];
       echo "<br>";



$i++;
       }
  if (!$result2) { 
     echo("Error, the query could not be executed: " .
     mysqli_error($db) . "</p>");
     mysqli_close($db);
     }

 $query1 = "SELECT state FROM states";
     $result1 = mysqli_query($db,$query1);


?>

Answer 1

您需要在循环外初始化$i，否则每次都会重置。而且你必须在每个echo之间递增它。

$i = 1;
while ($row = mysqli_fetch_assoc($result2)) {
    echo "var $i = {$row['model']}<br>";
    $i++;
    echo "var $i = {$row['price']}<br>";
    $i++;
}

Answer 2

我会

echo 'var data = '.json_encode($result2);

而不是

  while ($row = mysqli_fetch_assoc($result2)){
   $i=1;
   echo $database_model = $row['model'];
   echo "<br>";
   echo $database_price = $row['price'];
   echo "<br>";
   $i++;
   }

然后javascript变量data将包含结果数组

Answer 3

您可以动态创建变量：

$i=1;
while ($row = mysqli_fetch_assoc($result2)){
    ${'var' . $i++} = $row['model'];
    ${'var' . $i++} = $row['price'];
}

然后在你的脚本中你应该有：

echo $var1; // model #1
echo $var2; // price #1
echo $var3; // model #2
echo $var4; // price #2

我希望我会这样做：

$i=1;
while ($row = mysqli_fetch_assoc($result2)){
    ${'model' . $i} = $row['model'];
    ${'price' . $i++} = $row['price'];
}

这给了我：

echo $model1; // model #1
echo $price1; // price #1
echo $model2; // model #2
echo $price2; // price #2

如何分别存储数据库中的回显值？

3 个答案: