Question

我想将图像分配给PHP脚本中的变量，以便通过声明变量，可以在我想要的时候显示图像。

$FoodList = array_unique($FoodList);
if (!empty($FoodList)) {
    foreach ($FoodList as $key => $value) {
        // The variable would go here, so that image would appear
        //next to each variable
        echo "<li>" . $value . "<li>";
    }
    echo "</ul>";
}

Answer 1

你指定 $ var =“img src =”'your / pathto / image.ext'“;

$var = "your/pathto/image.ext";

并在html img代码中回显它

第二种方法更为优选

Answer 2

$FoodList = array_unique($FoodList);

if(!empty($FoodList)) {
    foreach ($FoodList as $key => $value) {
        //The variable would go here, so that image would appear
        //next to each variable
        $value = "<li>";

       //Maybe you'll only display an image is a certain condition is met? If so, then...
       if($condition == "parameter") {
           $value .= "<img src='path/to/img' alt='img' />";
       }

       $value .= "</li>";
       echo $value;
       unset($value);
    }
    echo "</ul>";
}

Answer 3

使用此：

echo "<li><img src='path_of_image/".$value."'/><li>";

假设$value的图片名称为图片扩展名。

Answer 4

$FoodList=array_unique($FoodList); 
$img_path = 'images/example.jpg';
if(!empty($FoodList)) 
{
   foreach ($FoodList as $key => $value)
{

   echo "<img src='$img_path' />";
   echo "<li>".$value."<li>";

}
echo "</ul>";
}

Answer 5

 <?php

     $name="Adil";
     echo $name;
     $path="FB_IMG_1465102989930.jpg";
     for($i=0;$i<44;$i++)
     {
       echo($i.'<br>')     ;
       if($i==10)
       {
           echo ".$path.";
           echo "<img src ='".$path."'>";
       }
     }
 ?>

Answer 6

请在图片名称前插入一个空格： - 示例： -

_collection.Aggregate()
   .Match(...)
   .Project(x => new { alias = x.IncludedField });

这里我在$image_name="myphoto.jpg"; $image_path="./upload/ ".$image_name;

之后添加了一个空格

将图像路径存储到MySql数据库中。

从您的HTML页面调用它： -

"./upload/(space)"

将图像分配给PHP变量

6 个答案: