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Php 从json对象获取图像url无效_Php_Json_Object - Fatal编程技术网

Php 从json对象获取图像url无效

Php 从json对象获取图像url无效,php,json,object,Php,Json,Object,我试图解码一个json对象,以便从中获取图像url。通常这很好,但不知怎么的,下面的代码不适合我 $partimages1 = $prods['images']; $partsimg1 = json_decode($partsimages1); if($partsimg1->image_intro != ''){ $productimages = '<img class="kw-prodimage-img" src="cms/'.$partsimg1->image_int

我试图解码一个json对象,以便从中获取图像url。通常这很好,但不知怎么的,下面的代码不适合我

$partimages1 = $prods['images'];
$partsimg1 = json_decode($partsimages1);

if($partsimg1->image_intro != ''){
  $productimages = '<img class="kw-prodimage-img" src="cms/'.$partsimg1->image_intro.'" alt="Product 1" title="Product 1" />';
}else{
  $productimages = '<img class="kw-prodimage-img" src="images/backgroundheader.jpg" alt="Product 1" title="Product 1" />';
}

出什么事了

您的打字错误已从
$partsimages1
更改为
$partimages1

$partsimg1 = json_decode($partimage1);

您的打字错误已从
$partsimages1
更改为
$partimages1

$partsimg1 = json_decode($partimage1);

有时需要一双新的眼睛才能注意到这样一个愚蠢的错误。。谢谢!有时需要一双新的眼睛才能注意到这样一个愚蠢的错误。。谢谢!打开你的错误报告。你会立即看到这个问题
error\u reporting=E\u ALL
打开错误报告。你会立即看到这个问题<代码>错误报告=E\u全部