如何在PHP

Question

我从我网站上生成的表单中构建了这个JSON。 我现在想将单个项目推送到数据库，所以我需要能够通过迭代来提取指令。

$ instruction0 $指令1

和坐在教学中的成分。

当尝试解析JSON时出现错误，我不知道如何在PHP中读取它。 （AKA PHP NOOB）我能够通过Python获得我所需的所有项目而没有任何问题。

Error

[
  [
    {
      "instruction": "Enter text here...asdada"
    },
    {
      "ingredient": "Beetroot",
      "amount": "2",
      "type": "grams"
    },
    {
      "ingredient": "Beetroot",
      "amount": "1",
      "type": "grams"
    }
  ],
  [
    {
      "instruction": "Enter text here..sdfsdf."
    },
    {
      "ingredient": "Carrot",
      "amount": "2",
      "type": "grams"
    },
    {
      "ingredient": "Beetroot",
      "amount": "525",
      "type": "grams"
    }
  ]
]

[ [ { "instruction": "Enter text here...asdada" }, { "ingredient": "Beetroot", "amount": "2", "type": "grams" }, { "ingredient": "Beetroot", "amount": "1", "type": "grams" } ], [ { "instruction": "Enter text here..sdfsdf." }, { "ingredient": "Carrot", "amount": "2", "type": "grams" }, { "ingredient": "Beetroot", "amount": "525", "type": "grams" } ] ]

Answer 1

您是否尝试使用第二个参数“true” json_decode 来指定您希望将输出作为关联数组。

json_decode($json, true);

其次，如果你手动将这个json字符串发布到你的代码中，你需要将它包装在引号内，所以基于那个

$json = '[ [ { "instruction": "Enter text here...asdada" }, { "ingredient": "Beetroot", "amount": "2", "type": "grams" }, { "ingredient": "Beetroot", "amount": "1", "type": "grams" } ], [ { "instruction": "Enter text here..sdfsdf." }, { "ingredient": "Carrot", "amount": "2", "type": "grams" }, { "ingredient": "Beetroot", "amount": "525", "type": "grams" } ] ] ';

$array = json_decode($json, true);

然后您可以正常迭代$array