Make share results distinguishable if there are more than one with the exact same display name
Signed-off-by: Julius Härtl <jus@bitgrid.net>
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@ -240,7 +240,27 @@ export default {
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// if there is a condition specified, filter it
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// if there is a condition specified, filter it
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const externalResults = this.externalResults.filter(result => !result.condition || result.condition(this))
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const externalResults = this.externalResults.filter(result => !result.condition || result.condition(this))
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this.suggestions = exactSuggestions.concat(suggestions).concat(externalResults).concat(lookupEntry)
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const allSuggestions = exactSuggestions.concat(suggestions).concat(externalResults).concat(lookupEntry)
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// Count occurances of display names in order to provide a distinguishable description if needed
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const nameCounts = allSuggestions.reduce((nameCounts, result) => {
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if (!result.displayName) {
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return nameCounts
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}
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if (!nameCounts[result.displayName]) {
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nameCounts[result.displayName] = 0
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}
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nameCounts[result.displayName]++
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return nameCounts
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}, {})
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this.suggestions = allSuggestions.map(item => {
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// Make sure that items with duplicate displayName get the shareWith applied as a description
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if (nameCounts[item.displayName] > 1 && !item.desc) {
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return { ...item, desc: item.shareWith }
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}
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return item
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})
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this.loading = false
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this.loading = false
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console.info('suggestions', this.suggestions)
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console.info('suggestions', this.suggestions)
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