"згодний" meaning in 烏克蘭語

See згодний in All languages combined, or Wiktionary

Adjective

IPA: [ˈzɦɔdnei̯] Audio: Uk-згодний.ogg Forms: згі́дний [alternative], зго́дний [canonical], zhódnyj [romanization]
Etymology: зго́да (zhóda) + -ний (-nyj)。
  1. 同意的
    Sense id: zh-згодний-uk-adj-45B9rPJ~
The following are not (yet) sense-disambiguated
{
  "categories": [
    {
      "kind": "other",
      "name": "含有後綴-ний的烏克蘭語詞",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "有1個詞條的頁面",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "有國際音標的烏克蘭語詞",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "有詞條的頁面",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "烏克蘭語形容詞",
      "parents": [],
      "source": "w"
    },
    {
      "kind": "other",
      "name": "烏克蘭語詞元",
      "parents": [],
      "source": "w"
    }
  ],
  "etymology_text": "зго́да (zhóda) + -ний (-nyj)。",
  "forms": [
    {
      "form": "згі́дний",
      "roman": "zhídnyj",
      "tags": [
        "alternative"
      ]
    },
    {
      "form": "зго́дний",
      "tags": [
        "canonical"
      ]
    },
    {
      "form": "zhódnyj",
      "tags": [
        "romanization"
      ]
    }
  ],
  "lang": "烏克蘭語",
  "lang_code": "uk",
  "pos": "adj",
  "pos_title": "形容詞",
  "senses": [
    {
      "glosses": [
        "同意的"
      ],
      "id": "zh-згодний-uk-adj-45B9rPJ~"
    }
  ],
  "sounds": [
    {
      "ipa": "[ˈzɦɔdnei̯]"
    },
    {
      "audio": "Uk-згодний.ogg",
      "mp3_url": "https://upload.wikimedia.org/wikipedia/commons/transcoded/6/62/Uk-згодний.ogg/Uk-згодний.ogg.mp3",
      "ogg_url": "https://commons.wikimedia.org/wiki/Special:FilePath/Uk-згодний.ogg",
      "raw_tags": [
        "音頻"
      ]
    }
  ],
  "word": "згодний"
}
{
  "categories": [
    "含有後綴-ний的烏克蘭語詞",
    "有1個詞條的頁面",
    "有國際音標的烏克蘭語詞",
    "有詞條的頁面",
    "烏克蘭語形容詞",
    "烏克蘭語詞元"
  ],
  "etymology_text": "зго́да (zhóda) + -ний (-nyj)。",
  "forms": [
    {
      "form": "згі́дний",
      "roman": "zhídnyj",
      "tags": [
        "alternative"
      ]
    },
    {
      "form": "зго́дний",
      "tags": [
        "canonical"
      ]
    },
    {
      "form": "zhódnyj",
      "tags": [
        "romanization"
      ]
    }
  ],
  "lang": "烏克蘭語",
  "lang_code": "uk",
  "pos": "adj",
  "pos_title": "形容詞",
  "senses": [
    {
      "glosses": [
        "同意的"
      ]
    }
  ],
  "sounds": [
    {
      "ipa": "[ˈzɦɔdnei̯]"
    },
    {
      "audio": "Uk-згодний.ogg",
      "mp3_url": "https://upload.wikimedia.org/wikipedia/commons/transcoded/6/62/Uk-згодний.ogg/Uk-згодний.ogg.mp3",
      "ogg_url": "https://commons.wikimedia.org/wiki/Special:FilePath/Uk-згодний.ogg",
      "raw_tags": [
        "音頻"
      ]
    }
  ],
  "word": "згодний"
}

Download raw JSONL data for згодний meaning in 烏克蘭語 (0.9kB)


This page is a part of the kaikki.org machine-readable 烏克蘭語 dictionary. This dictionary is based on structured data extracted on 2025-12-05 from the zhwiktionary dump dated 2025-12-01 using wiktextract (ddb1505 and 9905b1f). 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.