"plagia" meaning in 拉丁語

See plagia in All languages combined, or Wiktionary

Noun

Forms: plagiae [genitive]
Etymology: 可能是混合自 plaga (“区域”) 和古希臘語 πλάγια (plágia),源自古希臘語 πλάγιος (plágios,“倾斜”),對比希臘語 πλαγιά (plagiá,“山坡”)。
  1. (中世紀拉丁語) 斜坡
    Sense id: zh-plagia-la-noun-jz3vBIJC Categories (other): 中世紀拉丁語
  2. (中世紀拉丁語) 海灘
    Sense id: zh-plagia-la-noun-GPzOv44c Categories (other): 中世紀拉丁語
The following are not (yet) sense-disambiguated

Verb

Etymology: 請參閲主詞條的词源章節。
  1. plagiō 的第二人稱單數現在時主動態命令式 Tags: form-of Form of: plagiō
    Sense id: zh-plagia-la-verb-aC09QZvn
The following are not (yet) sense-disambiguated
{
  "categories": [
    {
      "kind": "other",
      "name": "拉丁語名詞",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "拉丁語第一類變格名詞",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "拉丁語詞元",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "拉丁語陰性名詞",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "拉丁語陰性名詞(第一類變格)",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "有6個詞條的頁面",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "有詞條的頁面",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "派生自古希臘語的意大利語詞",
      "parents": [],
      "source": "w"
    }
  ],
  "descendants": [
    {
      "lang": "達爾馬提亞語",
      "lang_code": "dlm",
      "word": "plui"
    },
    {
      "lang": "意大利語",
      "lang_code": "it",
      "roman": "plaga",
      "word": "piaggia"
    },
    {
      "descendants": [
        {
          "lang": "加利西亞語",
          "lang_code": "gl",
          "word": "praia"
        },
        {
          "lang": "葡萄牙語",
          "lang_code": "pt",
          "word": "praia"
        }
      ],
      "lang": "古葡萄牙語",
      "lang_code": "roa-opt",
      "word": "praya"
    },
    {
      "lang": "西西里語",
      "lang_code": "scn",
      "word": "praja"
    }
  ],
  "etymology_text": "可能是混合自 plaga (“区域”) 和古希臘語 πλάγια (plágia),源自古希臘語 πλάγιος (plágios,“倾斜”),對比希臘語 πλαγιά (plagiá,“山坡”)。",
  "forms": [
    {
      "form": "plagiae",
      "tags": [
        "genitive"
      ]
    }
  ],
  "lang": "拉丁語",
  "lang_code": "la",
  "pos": "noun",
  "pos_title": "名詞",
  "senses": [
    {
      "categories": [
        {
          "kind": "other",
          "name": "中世紀拉丁語",
          "parents": [],
          "source": "w"
        }
      ],
      "glosses": [
        "(中世紀拉丁語) 斜坡"
      ],
      "id": "zh-plagia-la-noun-jz3vBIJC"
    },
    {
      "categories": [
        {
          "kind": "other",
          "name": "中世紀拉丁語",
          "parents": [],
          "source": "w"
        }
      ],
      "glosses": [
        "(中世紀拉丁語) 海灘"
      ],
      "id": "zh-plagia-la-noun-GPzOv44c"
    }
  ],
  "tags": [
    "feminine"
  ],
  "word": "plagia"
}

{
  "categories": [
    {
      "kind": "other",
      "name": "拉丁語動詞變位形式",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "拉丁語非詞元形式",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "有6個詞條的頁面",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "有詞條的頁面",
      "parents": [],
      "source": "w"
    }
  ],
  "etymology_text": "請參閲主詞條的词源章節。",
  "lang": "拉丁語",
  "lang_code": "la",
  "pos": "verb",
  "pos_title": "動詞",
  "senses": [
    {
      "form_of": [
        {
          "word": "plagiō"
        }
      ],
      "glosses": [
        "plagiō 的第二人稱單數現在時主動態命令式"
      ],
      "id": "zh-plagia-la-verb-aC09QZvn",
      "tags": [
        "form-of"
      ]
    }
  ],
  "word": "plagia"
}
{
  "categories": [
    "拉丁語名詞",
    "拉丁語第一類變格名詞",
    "拉丁語詞元",
    "拉丁語陰性名詞",
    "拉丁語陰性名詞(第一類變格)",
    "有6個詞條的頁面",
    "有詞條的頁面",
    "派生自古希臘語的意大利語詞"
  ],
  "descendants": [
    {
      "lang": "達爾馬提亞語",
      "lang_code": "dlm",
      "word": "plui"
    },
    {
      "lang": "意大利語",
      "lang_code": "it",
      "roman": "plaga",
      "word": "piaggia"
    },
    {
      "descendants": [
        {
          "lang": "加利西亞語",
          "lang_code": "gl",
          "word": "praia"
        },
        {
          "lang": "葡萄牙語",
          "lang_code": "pt",
          "word": "praia"
        }
      ],
      "lang": "古葡萄牙語",
      "lang_code": "roa-opt",
      "word": "praya"
    },
    {
      "lang": "西西里語",
      "lang_code": "scn",
      "word": "praja"
    }
  ],
  "etymology_text": "可能是混合自 plaga (“区域”) 和古希臘語 πλάγια (plágia),源自古希臘語 πλάγιος (plágios,“倾斜”),對比希臘語 πλαγιά (plagiá,“山坡”)。",
  "forms": [
    {
      "form": "plagiae",
      "tags": [
        "genitive"
      ]
    }
  ],
  "lang": "拉丁語",
  "lang_code": "la",
  "pos": "noun",
  "pos_title": "名詞",
  "senses": [
    {
      "categories": [
        "中世紀拉丁語"
      ],
      "glosses": [
        "(中世紀拉丁語) 斜坡"
      ]
    },
    {
      "categories": [
        "中世紀拉丁語"
      ],
      "glosses": [
        "(中世紀拉丁語) 海灘"
      ]
    }
  ],
  "tags": [
    "feminine"
  ],
  "word": "plagia"
}

{
  "categories": [
    "拉丁語動詞變位形式",
    "拉丁語非詞元形式",
    "有6個詞條的頁面",
    "有詞條的頁面"
  ],
  "etymology_text": "請參閲主詞條的词源章節。",
  "lang": "拉丁語",
  "lang_code": "la",
  "pos": "verb",
  "pos_title": "動詞",
  "senses": [
    {
      "form_of": [
        {
          "word": "plagiō"
        }
      ],
      "glosses": [
        "plagiō 的第二人稱單數現在時主動態命令式"
      ],
      "tags": [
        "form-of"
      ]
    }
  ],
  "word": "plagia"
}

Download raw JSONL data for plagia meaning in 拉丁語 (1.6kB)


This page is a part of the kaikki.org machine-readable 拉丁語 dictionary. This dictionary is based on structured data extracted on 2025-06-26 from the zhwiktionary dump dated 2025-06-20 using wiktextract (074e7de and f1c2b61). The data shown on this site has been post-processed and various details (e.g., extra categories) removed, some information disambiguated, and additional data merged from other sources. See the raw data download page for the unprocessed wiktextract data.

If you use this data in academic research, please cite Tatu Ylonen: Wiktextract: Wiktionary as Machine-Readable Structured Data, Proceedings of the 13th Conference on Language Resources and Evaluation (LREC), pp. 1317-1325, Marseille, 20-25 June 2022. Linking to the relevant page(s) under https://kaikki.org would also be greatly appreciated.