revert showLink without the unused if-statement surrounding ignoreSendUrl

pull/5250/head
Ryo Hanafusa 2 months ago
parent 61c68e2305
commit d3c3808dae

@ -80,7 +80,7 @@ export default {
name: monitor.element.name, name: monitor.element.name,
monitor_index: monitor.index, monitor_index: monitor.index,
group_index: group.index, group_index: group.index,
isClickable: monitor.element.sendUrl, isClickable: this.showLink(monitor),
}; };
this.MonitorSettingDialog.show(); this.MonitorSettingDialog.show();
@ -95,6 +95,17 @@ export default {
toggleLink(groupIndex, index) { toggleLink(groupIndex, index) {
this.$root.publicGroupList[groupIndex].monitorList[index].sendUrl = !this.$root.publicGroupList[groupIndex].monitorList[index].sendUrl; this.$root.publicGroupList[groupIndex].monitorList[index].sendUrl = !this.$root.publicGroupList[groupIndex].monitorList[index].sendUrl;
}, },
/**
* Should a link to the monitor be shown?
* Attempts to guess if a link should be shown based upon if
* sendUrl is set and if the URL is default or not.
* @param {object} monitor Monitor to check
* @returns {boolean} Should the link be shown?
*/
showLink(monitor) {
return monitor.element.sendUrl && monitor.element.url && monitor.element.url !== "https://";
},
}, },
}; };
</script> </script>

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